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Sequence and Series — JEE Previous Year Questions

Every Sequence and Series question asked in JEE Main and JEE Advanced across the last 186 papers — 287 questions, each with its correct answer. Free to read, no account needed.

Questions

287

Papers it appeared in

164/186

Appearance rate

88%

All 287 Sequence and Series questions

Most recent papers first.

Q1·MathematicsSingle correctJEE Main 2026
Let α=3+4+8+9+13+14+…\alpha = 3 + 4 + 8 + 9 + 13 + 14 + \ldotsα=3+4+8+9+13+14+… upto 40 terms. If (tan⁡β)α1020(\tan \beta)^{\frac{\alpha}{1020}}(tanβ)1020α​ is a root of the equation x2+x−2=0x^2 + x - 2 = 0x2+x−2=0, β∈(0,π2)\beta \in \left(0, \frac{\pi}{2}\right)β∈(0,2π​), then sin⁡2β+3cos⁡2β\sin^2 \beta + 3\cos^2 \betasin2β+3cos2β is equal to:
  1. (A)2
  2. (B)74\frac{7}{4}47​
  3. (C)52\frac{5}{2}25​
  4. (D)32\frac{3}{2}23​

Correct answer: (A)

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
The value of 13^33 − 23^33 + 33^33 − ... + 153^33 is:
  1. (A)1706
  2. (B)1856
  3. (C)1982
  4. (D)2403

Correct answer: (B)

Step-by-step solution →
Q3·MathematicsNumericalJEE Main 2026
For the functions f(θ) = α tan2^22 θ + β cot2^22 θ, and g(θ) = α sin2^22 θ + β cos2^22 θ, α > β > 0, let min⁡0<θ<π/2f(θ)=max⁡0<θ<πg(θ)\min_{0<\theta<\pi/2} f(\theta) = \max_{0<\theta<\pi} g(\theta)min0<θ<π/2​f(θ)=max0<θ<π​g(θ). If the first term of a G.P. is (α2β)\left(\frac{\alpha}{2\beta}\right)(2βα​), its common ratio is (2βα)\left(\frac{2\beta}{\alpha}\right)(α2β​) and the sum of its first 10 terms is mn\frac{m}{n}nm​, gcd(m, n) = 1, then m + n is equal to ______.

Correct answer: 1279

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
The sum 1+12(12+22)+13(12+22+32)+…1 + \frac{1}{2}(1^{2} + 2^{2}) + \frac{1}{3}(1^{2} + 2^{2} + 3^{2}) + \ldots1+21​(12+22)+31​(12+22+32)+… upto 10 terms is equal to :
  1. (A)130
  2. (B)155
  3. (C)3152\frac{315}{2}2315​
  4. (D)3252\frac{325}{2}2325​

Correct answer: (C)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
Consider the quadratic equation (n2−2n+2)x2−3x+(n2−2n+2)2=0(n^{2} - 2n + 2)x^{2} - 3x + (n^{2} - 2n + 2)^{2} = 0(n2−2n+2)x2−3x+(n2−2n+2)2=0, n∈Rn \in \mathbb{R}n∈R. Let α be the minimum value of the product of its roots and β be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is α and the common ratio is αβ\frac{\alpha}{\beta}βα​, is :
  1. (A)6137\frac{61}{37}3761​
  2. (B)12181\frac{121}{81}81121​
  3. (C)364243\frac{364}{243}243364​
  4. (D)1093729\frac{1093}{729}7291093​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a G.P. is 8. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:
  1. (A)349\frac{34}{9}934​
  2. (B)3413\frac{34}{13}1334​
  3. (C)329\frac{32}{9}932​
  4. (D)3213\frac{32}{13}1332​

Correct answer: (A)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
∑n=110(528n(n+1)(n+2))\sum_{n=1}^{10}\left(\frac{528}{n(n+1)(n+2)}\right)∑n=110​(n(n+1)(n+2)528​) is equal to:
  1. (A)65
  2. (B)130
  3. (C)220
  4. (D)440

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
If the sum of the first 10 terms of the series 11+14×4+21+24×4+31+34×4+41+44×4+…\frac{1}{1 + 1^4 \times 4} + \frac{2}{1 + 2^4 \times 4} + \frac{3}{1 + 3^4 \times 4} + \frac{4}{1 + 4^4 \times 4} + \ldots1+14×41​+1+24×42​+1+34×43​+1+44×44​+… is mn\frac{m}{n}nm​, gcd(m, n) = 1, then m+nm + nm+n is equal to :
  1. (A)256
  2. (B)264
  3. (C)276
  4. (D)284

Correct answer: (C)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let A1,A2,A3,…,A39A_1, A_2, A_3, \ldots, A_{39}A1​,A2​,A3​,…,A39​ be 39 arithmetic means between the numbers 59 and 159. Then the mean of A25,A28,A31A_{25}, A_{28}, A_{31}A25​,A28​,A31​ and A36A_{36}A36​ is equal to :
  1. (A)129
  2. (B)136
  3. (C)131.50
  4. (D)134

Correct answer: (D)

Step-by-step solution →
Q10·MathematicsSingle correctJEE Main 2026
Let the sum of the first nnn terms of an A.P. be 3n2+5n3n^2 + 5n3n2+5n. Then the sum of squares of the first 10 terms of the A.P. is:
  1. (A)10220
  2. (B)12860
  3. (C)15220
  4. (D)19780

Correct answer: (C)

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
The first term of an A.P. of 30 non-negative terms is 103\frac{10}{3}310​. If the sum of this A.P. is the cube of its last term, then its common difference is:
  1. (A)587\frac{5}{87}875​
  2. (B)2583\frac{25}{83}8325​
  3. (C)1529\frac{15}{29}2915​
  4. (D)529\frac{5}{29}295​

Correct answer: (A)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let α=14+18+116+…∞\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \inftyα=41​+81​+161​+…∞ and β=13+19+127+…∞\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \inftyβ=31​+91​+271​+…∞. Then the value of (0.2)log⁡5(α)+(0.04)log⁡5(β)(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}(0.2)log5​​(α)+(0.04)log5​(β) is equal to:
  1. (A)4
  2. (B)5
  3. (C)8
  4. (D)25

Correct answer: (C)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be an A.P. and g1=a1,g2,g3,…g_1 = a_1, g_2, g_3, \ldotsg1​=a1​,g2​,g3​,… be an increasing G.P. If a1=a2+g2=1a_1 = a_2 + g_2 = 1a1​=a2​+g2​=1 and a3+g3=4a_3 + g_3 = 4a3​+g3​=4, then a10+g5a_{10} + g_5a10​+g5​ is equal to:
  1. (A)818181
  2. (B)767676
  3. (C)626262
  4. (D)555555

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A ∩ B, which are divisible by 3, is :
  1. (A)4
  2. (B)5
  3. (C)6
  4. (D)7

Correct answer: (B)

Step-by-step solution →
Q15·MathematicsNumericalJEE Main 2026
If ∑k=1nak=6n3\sum_{k=1}^{n} a_{k} = 6n^{3}∑k=1n​ak​=6n3, then ∑k=16(ak+1−ak36)2\sum_{k=1}^{6}\left(\frac{a_{k+1} - a_{k}}{36}\right)^{2}∑k=16​(36ak+1​−ak​​)2 is equal to ________.

Correct answer: 91

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2026
The sum 131+13+231+3+13+23+331+3+5+⋯\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \cdots113​+1+313+23​+1+3+513+23+33​+⋯ up to 8 terms, is:
  1. (A)707070
  2. (B)717171
  3. (C)727272
  4. (D)737373

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
The common difference of the A.P.: a1a_1a1​, a2a_2a2​, …., ama_mam​ is 13 more than the common difference of the A.P.: b1b_1b1​, b2b_2b2​, …., bnb_nbn​. If b31b_{31}b31​ = –277, b43b_{43}b43​ = –385 and a78a_{78}a78​ = 327, then a1a_1a1​ is equal to
  1. (A)21
  2. (B)24
  3. (C)19
  4. (D)16

Correct answer: (C)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
The value of ∑k=1∞(−1)k+1(k(k+1)k!)\sum\limits_{k=1}^{\infty} (-1)^{k+1}\left(\dfrac{k(k+1)}{k!}\right)k=1∑∞​(−1)k+1(k!k(k+1)​) is :
  1. (A)2/e2/e2/e
  2. (B)1/e1/e1/e
  3. (C)e\sqrt{e}e​
  4. (D)e/2e/2e/2

Correct answer: (B)

Step-by-step solution →
Q19·MathematicsNumericalJEE Main 2026
In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is R\mathbb{R}R – (a, b ), then a2+b2a^{2} + b^{2}a2+b2 is equal to ____.

Correct answer: 90

Step-by-step solution →
Q20·MathematicsNumericalJEE Main 2026
If ∑r=125(rr4+r2+1)=pq\sum_{r=1}^{25}\left(\frac{r}{r^4 + r^2 + 1}\right) = \frac{p}{q}∑r=125​(r4+r2+1r​)=qp​, where p and q are positive integers such that gcd (p, q) = 1, then p + q is equal to ______.

Correct answer: 976

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2026
6326+10.1325+10.2324+10.22323+...+10.2243\frac{6}{3^{26}} + \frac{10.1}{3^{25}} + \frac{10.2}{3^{24}} + \frac{10.2^2}{3^{23}} + ...+ \frac{10.2^{24}}{3}3266​+32510.1​+32410.2​+32310.22​+...+310.224​ is equal to
  1. (A)2252^{25}225
  2. (B)2262^{26}226
  3. (C)3253^{25}325
  4. (D)3263^{26}326

Correct answer: (B)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2026
Let 729, 81, 9, 1, …. be a sequence and PnP_{n}Pn​ denote the product of the first n terms of this sequence. If 2∑n=140(Pn)1n=3α−13β2\sum_{n=1}^{40}\left(P_{n}\right)^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}2∑n=140​(Pn​)n1​=3β3α−1​ and gcd (α, β) = 1, then α + β is equal to
  1. (A)737373
  2. (B)747474
  3. (C)757575
  4. (D)767676

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2026
Consider an A.P.: a1,a2,....,ana_{1}, a_{2},....,a_{n}a1​,a2​,....,an​; a1>0a_{1} > 0a1​>0. If a2−a1=−34a_{2} - a_{1} = \frac{-3}{4}a2​−a1​=4−3​, an=14a1a_{n} = \frac{1}{4} a_{1}an​=41​a1​, and ∑i=1nai=5252\sum\limits_{i=1}^{n} a_{i} = \frac{525}{2}i=1∑n​ai​=2525​, then ∑i=117ai\sum\limits_{i=1}^{17} a_{i}i=1∑17​ai​ is equal to
  1. (A)476
  2. (B)952
  3. (C)238
  4. (D)136

Correct answer: (C)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2026
(13+47)+(132+13×47+4272)+(133+132×47+13×4272+4373)\left(\frac{1}{3} + \frac{4}{7}\right) + \left(\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}\right) + \left(\frac{1}{3^{3}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}\right)(31​+74​)+(321​+31​×74​+7242​)+(331​+321​×74​+31​×7242​+7343​) + ...... upto infinite terms is equal to -
  1. (A)52\frac{5}{2}25​
  2. (B)74\frac{7}{4}47​
  3. (C)43\frac{4}{3}34​
  4. (D)65\frac{6}{5}56​

Correct answer: (A)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2026
Let ∑k=1nak=αn2+βn\sum_{k=1}^{n} a_k = \alpha n^2 + \beta n∑k=1n​ak​=αn2+βn. If a10=59a_{10} = 59a10​=59 and a6=7a1a_6 = 7a_1a6​=7a1​ then α + β is equal to
  1. (A)12
  2. (B)3
  3. (C)5
  4. (D)7

Correct answer: (C)

Step-by-step solution →
Q26·MathematicsNumericalJEE Main 2026
Suppose a, b, c are in A.P. and a2a^2a2, 2b22b^22b2, c2c^2c2 are in G.P. If a<b<ca < b < ca<b<c and a+b+c=1a + b + c = 1a+b+c=1, then 9(a2+b2+c2)9(a^2 + b^2 + c^2)9(a2+b2+c2) is equal to______ .

Correct answer: 9

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2026
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
  1. (A)–20
  2. (B)–24
  3. (C)–26
  4. (D)–22

Correct answer: (D)

Step-by-step solution →
Q28·MathematicsSingle correctJEE Main 2026
Let a1,a22,a322,…,a1029a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}a1​,2a2​​,22a3​​,…,29a10​​ be a G.P. of common ratio 12\frac{1}{\sqrt{2}}2​1​. If a1+a2+…+a10=62a_1 + a_2 + \ldots + a_{10} = 62a1​+a2​+…+a10​=62, then a1a_1a1​ is equal to :
  1. (A)2(2−1)2\left(\sqrt{2} - 1\right)2(2​−1)
  2. (B)2−22 - \sqrt{2}2−2​
  3. (C)2−1\sqrt{2} - 12​−1
  4. (D)2(2−2)2(2 - \sqrt{2})2(2−2​)

Correct answer: (A)

Step-by-step solution →
Q29·MathematicsNumericalJEE Main 2026
Let a1=1a_{1}=1a1​=1and for n≥1n\geq 1n≥1, an+1=12an+n2−2n−1n2(n+1)2a_{n+1}=\frac{1}{2}a_{n}+\frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}an+1​=21​an​+n2(n+1)2n2−2n−1​. Then ∣∑n=1∞(an−2n2)∣\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|​∑n=1∞​(an​−n22​)​ is equal to ________.

Correct answer: 2

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2026
Let a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​,….. be a G.P. of increasing positive terms such that a2.a3.a4=64a_{2}.a_{3}.a_{4} = 64a2​.a3​.a4​=64 and a1+a3+a5=8137a_{1} + a_{3} + a_{5} = \frac{813}{7}a1​+a3​+a5​=7813​ . Then a3+a5+a7a_{3} + a_{5} + a_{7}a3​+a5​+a7​ is equal to :
  1. (A)3256
  2. (B)3252
  3. (C)3244
  4. (D)3248

Correct answer: (B)

Step-by-step solution →
Q31·MathematicsNumericalJEE Advanced 2025
Let R denote the set of all real numbers. Let f:R→Rf: R \to Rf:R→R be a function such that f(x)>0f(x) > 0f(x)>0 for all x∈Rx \in Rx∈R, and f(x+y)=f(x)f(y)f(x + y) = f(x) f(y)f(x+y)=f(x)f(y) for all x,y∈Rx, y \in Rx,y∈R. Let the real numbers a1a_1a1​, a2a_2a2​, ....., a50a_{50}a50​ be in an arithmetic progression. If f(a31)=64f(a25)f(a_{31}) = 64 f(a_{25})f(a31​)=64f(a25​), and ∑i=150f(ai)=3(225+1)\sum_{i=1}^{50} f(a_i) = 3\left(2^{25} + 1\right)∑i=150​f(ai​)=3(225+1), then the value of ∑i=630f(ai)\sum_{i=6}^{30} f(a_i)∑i=630​f(ai​) is ________ .

Correct answer: 96

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
If 114+124+134+…∞=π490\dfrac{1}{1^4}+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\ldots\infty=\dfrac{\pi^4}{90}141​+241​+341​+…∞=90π4​, 114+134+154+…∞=α\dfrac{1}{1^4}+\dfrac{1}{3^4}+\dfrac{1}{5^4}+\ldots\infty=\alpha141​+341​+541​+…∞=α, 124+144+164+…∞=β\dfrac{1}{2^4}+\dfrac{1}{4^4}+\dfrac{1}{6^4}+\ldots\infty=\beta241​+441​+641​+…∞=β, then αβ\dfrac{\alpha}{\beta}βα​ is equal to:
  1. (A)232323
  2. (B)181818
  3. (C)151515
  4. (D)141414

Correct answer: (C)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
Let ana_nan​ be the nthn^{th}nth term of an A.P. If Sn=a1+a2+a3+⋯+an=700S_n=a_1+a_2+a_3+\dots+a_n=700Sn​=a1​+a2​+a3​+⋯+an​=700, a6=7a_6=7a6​=7 and S7=7S_7=7S7​=7, then ana_nan​ is equal to:
  1. (A)56
  2. (B)65
  3. (C)64
  4. (D)70

Correct answer: (C)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:
  1. (A)760
  2. (B)755
  3. (C)750
  4. (D)757

Correct answer: (D)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
Let x1,x2,x3,x4x_1,x_2,x_3,x_4x1​,x2​,x3​,x4​ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1,x2,x3,x4x_1,x_2,x_3,x_4x1​,x2​,x3​,x4​ then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4)\dfrac{1}{24}(x_1x_2x_3x_4)241​(x1​x2​x3​x4​) is:
  1. (A)72
  2. (B)18
  3. (C)36
  4. (D)216

Correct answer: (D)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Consider two sets AAA and BBB, each containing three numbers in A.P. Let the sum and the product of the elements of AAA be 36 and ppp respectively and the sum and the product of the elements of BBB be 36 and qqq respectively. Let ddd and DDD be the common differences of the APs in AAA and BBB respectively such that D=d+3D=d+3D=d+3, d>0d>0d>0. If p+qp−q=195\dfrac{p+q}{p-q}=\dfrac{19}{5}p−qp+q​=519​, then p−qp-qp−q is equal to:
  1. (A)600
  2. (B)450
  3. (C)630
  4. (D)540

Correct answer: (D)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
Let A={1,6,11,16,…}A=\{1,6,11,16,\ldots\}A={1,6,11,16,…} and B={9,16,23,30,…}B=\{9,16,23,30,\ldots\}B={9,16,23,30,…} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B)n(A\cup B)n(A∪B) is
  1. (A)3814
  2. (B)4027
  3. (C)3761
  4. (D)4003

Correct answer: (C)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
1+3+52+7+92+…1+3+5^2+7+9^2+\ldots1+3+52+7+92+… upto 40 terms is equal to
  1. (A)43890
  2. (B)41880
  3. (C)33980
  4. (D)40870

Correct answer: (B)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
If the sum of the first 20 terms of the series 4⋅14+3⋅12+14+4⋅24+3⋅22+24+4⋅34+3⋅32+34+…\dfrac{4\cdot1}{4+3\cdot1^2+1^4}+\dfrac{4\cdot2}{4+3\cdot2^2+2^4}+\dfrac{4\cdot3}{4+3\cdot3^2+3^4}+\dots4+3⋅12+144⋅1​+4+3⋅22+244⋅2​+4+3⋅32+344⋅3​+… is mn\dfrac{m}{n}nm​, where mmm and nnn are coprime, then m+nm+nm+n is equal to:
  1. (A)423
  2. (B)420
  3. (C)421
  4. (D)422

Correct answer: (C)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
The sum 1+3+11+25+45+71+…1+3+11+25+45+71+\ldots1+3+11+25+45+71+… upto 20 terms, is equal to:
  1. (A)7240
  2. (B)7130
  3. (C)6982
  4. (D)8124

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
The sum 1+1+32!+1+3+53!+1+3+5+74!+…1+\dfrac{1+3}{2!}+\dfrac{1+3+5}{3!}+\dfrac{1+3+5+7}{4!}+\ldots1+2!1+3​+3!1+3+5​+4!1+3+5+7​+… upto ∞\infty∞ terms, is equal to:
  1. (A)6e6e6e
  2. (B)4e4e4e
  3. (C)3e3e3e
  4. (D)2e2e2e

Correct answer: (D)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Let a1,a2,a3,…a_1,a_2,a_3,\ldotsa1​,a2​,a3​,… be a G.P. of increasing positive numbers. If a3a5=729a_3a_5=729a3​a5​=729 and a2+a4=1114a_2+a_4=\dfrac{111}{4}a2​+a4​=4111​, then 24(a1+a2+a3)24(a_1+a_2+a_3)24(a1​+a2​+a3​) is equal to:
  1. (A)131
  2. (B)130
  3. (C)129
  4. (D)128

Correct answer: (C)

Step-by-step solution →
Q43·MathematicsIntegerJEE Main 2025
If the sum of the first 10 terms of the series 4⋅11+4⋅14+4⋅21+4⋅24+4⋅31+4⋅34+…\dfrac{4\cdot1}{1+4\cdot1^4}+\dfrac{4\cdot2}{1+4\cdot2^4}+\dfrac{4\cdot3}{1+4\cdot3^4}+\dots1+4⋅144⋅1​+1+4⋅244⋅2​+1+4⋅344⋅3​+… is mn\dfrac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is equal to ______.

Correct answer: 441

Step-by-step solution →
Q44·MathematicsSingle correctJEE Main 2025
Let a1,a2,a3,…a_1, a_2, a_3,\ldotsa1​,a2​,a3​,… be in an A.P. such that ∑k=112a2k−1=−725a1\displaystyle\sum_{k=1}^{12}a_{2k-1}=-\dfrac{72}{5}a_1k=1∑12​a2k−1​=−572​a1​, a1≠0a_1\ne0a1​=0. If ∑k=1nak=0\displaystyle\sum_{k=1}^{n}a_k=0k=1∑n​ak​=0, then nnn is:
  1. (A)11
  2. (B)10
  3. (C)18
  4. (D)17

Correct answer: (A)

Step-by-step solution →
Q45·MathematicsSingle correctJEE Main 2025
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 212\dfrac{21}{2}221​. Then the number of terms which are integers in the A.P. is:
  1. (A)4
  2. (B)10
  3. (C)6
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q46·MathematicsIntegerJEE Main 2025
Let a1,a2,…,a2024a_1, a_2,\ldots,a_{2024}a1​,a2​,…,a2024​ be an Arithmetic Progression such that a1+(a5+a10+a15+⋯+a2020)+a2024=2233a_1+(a_5+a_{10}+a_{15}+\cdots+a_{2020})+a_{2024}=2233a1​+(a5​+a10​+a15​+⋯+a2020​)+a2024​=2233. Then a1+a2+a3+⋯+a2024a_1+a_2+a_3+\cdots+a_{2024}a1​+a2​+a3​+⋯+a2024​ is equal to ______.

Correct answer: 11132

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2025
Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th11^{\text{th}}11th term is:
  1. (A)84
  2. (B)122
  3. (C)90
  4. (D)108

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2025
The value of lim⁡n→∞(∑k=1nk3+6k2+11k+5(k+3)!)\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^{n}\dfrac{k^3+6k^2+11k+5}{(k+3)!}\right)n→∞lim​(k=1∑n​(k+3)!k3+6k2+11k+5​) is:
  1. (A)43\dfrac{4}{3}34​
  2. (B)222
  3. (C)73\dfrac{7}{3}37​
  4. (D)53\dfrac{5}{3}35​

Correct answer: (D)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2025
Let TrT_rTr​ be the rthr^{th}rth term of an A.P. If for some m, Tm=125T_m=\frac{1}{25}Tm​=251​, T25=120T_{25}=\frac{1}{20}T25​=201​ and 20∑r=125Tr=1320\sum_{r=1}^{25} T_r=1320∑r=125​Tr​=13, then 5m∑r=m2mTr5m\sum_{r=m}^{2m} T_r5m∑r=m2m​Tr​ is equal to:
  1. (A)112
  2. (B)126
  3. (C)98
  4. (D)142

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2025
Let ⟨an⟩\langle a_n\rangle⟨an​⟩ be a sequence such that a0=0a_0=0a0​=0, a1=12a_1=\frac{1}{2}a1​=21​ and 2an+2=5an+1−3an2a_{n+2}=5a_{n+1}-3a_n2an+2​=5an+1​−3an​, n=0,1,2,3,…n=0, 1, 2, 3, \ldotsn=0,1,2,3,… Then ∑k=1100ak\sum_{k=1}^{100} a_k∑k=1100​ak​ is equal to:
  1. (A)3a99−1003a_{99}-1003a99​−100
  2. (B)3a100−1003a_{100}-1003a100​−100
  3. (C)3a100+1003a_{100}+1003a100​+100
  4. (D)3a99+1003a_{99}+1003a99​+100

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2025
For positive integers n, if 4an=(n2+5n+6)4a_n=(n^2+5n+6)4an​=(n2+5n+6) and Sn=∑k=1n1akS_n=\sum_{k=1}^{n}\frac{1}{a_k}Sn​=∑k=1n​ak​1​, then the value of 507 S2025507\,S_{2025}507S2025​ is:
  1. (A)540
  2. (B)1350
  3. (C)675
  4. (D)135

Correct answer: (C)

Step-by-step solution →
Q52·MathematicsIntegerJEE Main 2025
The interior angles of a polygon with n sides, are in an A.P. with common difference 6∘6^\circ6∘. If the largest interior angle of the polygon is 219∘219^\circ219∘, then n is equal to ______.

Correct answer: 20

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2025
In an arithmetic progression, if S40=1030S_{40}=1030S40​=1030 and S12=57S_{12}=57S12​=57, then S30−S10S_{30}-S_{10}S30​−S10​ is equal to:
  1. (A)510
  2. (B)515
  3. (C)525
  4. (D)505

Correct answer: (B)

Step-by-step solution →
Q54·MathematicsSingle correctJEE Main 2025
Let Sn=12+16+112+120+…S_n=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\ldotsSn​=21​+61​+121​+201​+… upto n terms. If the sum of the first six terms of an A.P. with first term −p-p−p and common difference ppp is 2026 S2025\sqrt{2026\,S_{2025}}2026S2025​​, then the absolute difference between 20th20^{th}20th and 15th15^{th}15th terms of the A.P. is
  1. (A)252525
  2. (B)909090
  3. (C)202020
  4. (D)454545

Correct answer: (A)

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2025
If 7=5+17(5+α)+172(5+2α)+173(5+3α)+…∞7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\dfrac{1}{7^3}(5+3\alpha)+\ldots\infty7=5+71​(5+α)+721​(5+2α)+731​(5+3α)+…∞, then the value of α\alphaα is:
  1. (A)1
  2. (B)67\dfrac{6}{7}76​
  3. (C)6
  4. (D)17\dfrac{1}{7}71​

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2025
If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to
  1. (A)−1200-1200−1200
  2. (B)−1080-1080−1080
  3. (C)−1020-1020−1020
  4. (D)−120-120−120

Correct answer: (B)

Step-by-step solution →
Q57·MathematicsIntegerJEE Main 2025
The roots of the quadratic equation 3x2−px+q=03x^2-px+q=03x2−px+q=0 are 10th10^{\text{th}}10th and 11th11^{\text{th}}11th terms of an arithmetic progression with common difference 32\dfrac{3}{2}23​. If the sum of the first 111111 terms of this arithmetic progression is 888888, then q−2pq-2pq−2p is equal to __________.

Correct answer: 474

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2025
Let a1,a2,…a_1,a_2,\dotsa1​,a2​,… be a G.P. of increasing positive terms. If a1a5=28a_1a_5=28a1​a5​=28 and a2+a4=29a_2+a_4=29a2​+a4​=29, then a6a_6a6​ equals:
  1. (A)628
  2. (B)526
  3. (C)784
  4. (D)812

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsIntegerJEE Main 2025
A finite sum of the form ∑(r+2)/(2r+1)\sum (r+2)/(2^{r+1})∑(r+2)/(2r+1)-type series evaluates to mn\tfrac{m}{n}nm​ with gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1. Find m−nm-nm−n.

Correct answer: 2035

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2025
Suppose that the number of terms in an A.P. is 2k2k2k, k∈Nk\in\mathbb{N}k∈N. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then kkk is equal to
  1. (A)5
  2. (B)8
  3. (C)6
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2025
If ∑r=1nTr=(2n−1)(2n+1)(2n+3)(2n+5)64\displaystyle\sum_{r=1}^{n}T_r=\dfrac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}r=1∑n​Tr​=64(2n−1)(2n+1)(2n+3)(2n+5)​, then lim⁡n→∞∑r=1n1Tr\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\dfrac1{T_r}n→∞lim​r=1∑n​Tr​1​ equals:
  1. (A)1
  2. (B)0
  3. (C)23\tfrac2332​
  4. (D)13\tfrac1331​

Correct answer: (C)

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2024
If a function fff satisfies f(m+n)=f(m)+f(n)f(m + n) = f(m) + f(n)f(m+n)=f(m)+f(n) for all m,n∈Nm, n \in \mathbb{N}m,n∈N and f(1)=1f(1) = 1f(1)=1, then the largest natural number λ\lambdaλ such that ∑k=12022f(λ+k)≤(2022)2\sum_{k=1}^{2022} f(\lambda + k) \le (2022)^2∑k=12022​f(λ+k)≤(2022)2 is equal to ________.

Correct answer: 1010

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2024
Let aaa, ararar, ar2ar^{2}ar2, ... be an infinite G.P. If ∑n=0∞arn=57\displaystyle\sum_{n=0}^{\infty}ar^{n}=57n=0∑∞​arn=57 and ∑n=0∞a3r3n=9747\displaystyle\sum_{n=0}^{\infty}a^{3}r^{3n}=9747n=0∑∞​a3r3n=9747, then a+18ra+18ra+18r is equal to:
  1. (A)272727
  2. (B)464646
  3. (C)383838
  4. (D)313131

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
If the sum of series 11⋅(1+d)+1(1+d)(1+2d)+…+1(1+9d)(1+10d)\frac{1}{1\cdot(1 + d)} + \frac{1}{(1 + d)(1 + 2d)} + \ldots + \frac{1}{(1 + 9d)(1 + 10d)}1⋅(1+d)1​+(1+d)(1+2d)1​+…+(1+9d)(1+10d)1​ is equal to 5, then 50d50d50d is equal to:
  1. (A)20
  2. (B)5
  3. (C)15
  4. (D)10

Correct answer: (B)

Step-by-step solution →
Q65·MathematicsNumericalJEE Main 2024
If (1α+1+1α+2+⋯+1α+1012)−(12⋅1+14⋅3+16⋅5+⋯+12024⋅2023)=12024\left(\dfrac{1}{\alpha+1}+\dfrac{1}{\alpha+2}+\cdots+\dfrac{1}{\alpha+1012}\right)-\left(\dfrac{1}{2\cdot 1}+\dfrac{1}{4\cdot 3}+\dfrac{1}{6\cdot 5}+\cdots+\dfrac{1}{2024\cdot 2023}\right)=\dfrac{1}{2024}(α+11​+α+21​+⋯+α+10121​)−(2⋅11​+4⋅31​+6⋅51​+⋯+2024⋅20231​)=20241​, then α\alphaα is equal to _______.

Correct answer: 1011

Step-by-step solution →
Q66·MathematicsNumericalJEE Main 2024
If the range of f(θ)=sin⁡4θ+3cos⁡2θsin⁡4θ+cos⁡2θf(\theta)=\dfrac{\sin^4\theta+3\cos^2\theta}{\sin^4\theta+\cos^2\theta}f(θ)=sin4θ+cos2θsin4θ+3cos2θ​, θ∈R\theta\in\mathbb{R}θ∈R is [α,β][\alpha,\beta][α,β], then the sum of the infinite G.P., whose first term is 646464 and the common ratio is αβ\dfrac\alpha\betaβα​, is equal to ___

Correct answer: 96

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 703\dfrac{70}{3}370​ and the product of the third and fifth terms is 494949. Then the sum of the 444th, 666th and 888th terms is
  1. (A)969696
  2. (B)787878
  3. (C)919191
  4. (D)848484

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsNumericalJEE Main 2024
An arithmetic progression is written in the following way: row 1 is 222; row 2 is 5, 85,\ 85, 8; row 3 is 11, 14, 1711,\ 14,\ 1711, 14, 17; row 4 is 20, 23, 26, 2920,\ 23,\ 26,\ 2920, 23, 26, 29; and so on. The sum of all the terms of the 101010th row is _______ .

Correct answer: 1505

Step-by-step solution →
Q69·MathematicsNumericalJEE Main 2024
Let the positive integers be written in the form: 111 (row 1); 2,32,32,3 (row 2); 4,5,64,5,64,5,6 (row 3); 7,8,9,107,8,9,107,8,9,10 (row 4); and so on. If the kthk^{th}kth row contains exactly kkk numbers for every natural number kkk, then the row in which the number 531053105310 will be, is ___

Correct answer: 103

Step-by-step solution →
Q70·MathematicsNumericalJEE Main 2024
Let the first term of a series be T1=6T_{1}=6T1​=6 and its rthr^{\text{th}}rth term Tr=3Tr−1+6rT_{r}=3T_{r-1}+6^{r}Tr​=3Tr−1​+6r, r=2,3,…,nr=2,3,\ldots,nr=2,3,…,n. If the sum of the first nnn terms of this series is 15(n2−12n+39)(4⋅6n−5⋅3n+1)\dfrac{1}{5}(n^{2}-12n+39)(4\cdot 6^{n}-5\cdot 3^{n}+1)51​(n2−12n+39)(4⋅6n−5⋅3n+1), then nnn is equal to _______.

Correct answer: 6

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2024
A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of m is equal to:
  1. (A)125
  2. (B)150
  3. (C)180
  4. (D)160

Correct answer: (B)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2024
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then:
  1. (A)P2=363 QP^2 = 36\sqrt{3}\,QP2=363​Q
  2. (B)P2=63 QP^2 = 6\sqrt{3}\,QP2=63​Q
  3. (C)P=363 Q2P = 36\sqrt{3}\,Q^2P=363​Q2
  4. (D)P2=723 QP^2 = 72\sqrt{3}\,QP2=723​Q

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsNumericalJEE Main 2024
If 1+3−223+5−2618+93−112363+49−206180+⋯1+\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}+\dfrac{5-2\sqrt{6}}{18}+\dfrac{9\sqrt{3}-11\sqrt{2}}{36\sqrt{3}}+\dfrac{49-20\sqrt{6}}{180}+\cdots1+23​3​−2​​+185−26​​+363​93​−112​​+18049−206​​+⋯ upto ∞=2(ba+1)log⁡e(ab)\infty=2\left(\sqrt{\dfrac{b}{a}}+1\right)\log_e\left(\dfrac{a}{b}\right)∞=2(ab​​+1)loge​(ba​), where aaa and bbb are integers with gcd⁡(a,b)=1\gcd(a,b)=1gcd(a,b)=1, then 11a+18b11a+18b11a+18b is equal to __________.

Correct answer: 76

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2024
If 11+2+12+3+…+199+100=m\dfrac{1}{\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}} + \ldots + \dfrac{1}{\sqrt{99}+\sqrt{100}} = m1​+2​1​+2​+3​1​+…+99​+100​1​=m and 11⋅2+12⋅3+…+199⋅100=n\dfrac{1}{1\cdot 2} + \dfrac{1}{2\cdot 3} + \ldots + \dfrac{1}{99\cdot 100} = n1⋅21​+2⋅31​+…+99⋅1001​=n, then the point (m,n)(m, n)(m,n) lies on the line:
  1. (A)11(x−1)−100(y−2)=011(x - 1) - 100(y - 2) = 011(x−1)−100(y−2)=0
  2. (B)11(x−2)−100(y−1)=011(x - 2) - 100(y - 1) = 011(x−2)−100(y−1)=0
  3. (C)11(x−1)−100y=011(x - 1) - 100y = 011(x−1)−100y=0
  4. (D)11x−100y=011x - 100y = 011x−100y=0

Correct answer: (D)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2024
For x≥0x\ge 0x≥0, the least value of KKK for which 41+x+41−x4^{1+x}+4^{1-x}41+x+41−x, K2\dfrac{K}{2}2K​, 16x+16−x16^x+16^{-x}16x+16−x are three consecutive terms of an A.P. is equal to:
  1. (A)10
  2. (B)4
  3. (C)8
  4. (D)16

Correct answer: (A)

Step-by-step solution →
Q76·MathematicsNumericalJEE Main 2024
Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be in an arithmetic progression of positive terms. Let Ak=a12−a22+a32−a42+…+a2k−12−a2k2A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2k-1}^2 - a_{2k}^2Ak​=a12​−a22​+a32​−a42​+…+a2k−12​−a2k2​. If A3=−153A_3 = -153A3​=−153, A5=−435A_5 = -435A5​=−435 and a12+a22+a32=66a_1^2 + a_2^2 + a_3^2 = 66a12​+a22​+a32​=66, then a17−A7a_{17} - A_7a17​−A7​ is equal to ___.

Correct answer: 910

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2024
Let the first three terms 222, ppp and qqq, with q≠2q\ne2q=2, of a G.P. be respectively the 7th7^{th}7th, 8th8^{th}8th and 13th13^{th}13th terms of an A.P. If the 5th5^{th}5th term of the G.P. is the nthn^{th}nth term of the A.P., then nnn is equal to:
  1. (A)151151151
  2. (B)169169169
  3. (C)177177177
  4. (D)163163163

Correct answer: (D)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2024
The value of 1⋅22+2⋅32+…+100⋅(101)212⋅2+22⋅3+…+1002⋅101\dfrac{1\cdot 2^2+2\cdot 3^2+\ldots+100\cdot(101)^2}{1^2\cdot 2+2^2\cdot 3+\ldots+100^2\cdot 101}12⋅2+22⋅3+…+1002⋅1011⋅22+2⋅32+…+100⋅(101)2​ is
  1. (A)306305\tfrac{306}{305}305306​
  2. (B)305301\tfrac{305}{301}301305​
  3. (C)3231\tfrac{32}{31}3132​
  4. (D)3130\tfrac{31}{30}3031​

Correct answer: (B)

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2024
Let three real numbers aaa, bbb, ccc be in arithmetic progression and a+1a+1a+1, bbb, c+3c+3c+3 be in geometric progression. If a>10a>10a>10 and the arithmetic mean of aaa, bbb and ccc is 8, then the cube of the geometric mean of aaa, bbb and ccc is
  1. (A)120
  2. (B)312
  3. (C)316
  4. (D)128

Correct answer: (A)

Step-by-step solution →
Q80·MathematicsNumericalJEE Main 2024
Let a=1+2C23!+3C24!+4C25!+…a=1+\dfrac{{}^2C_2}{3!}+\dfrac{{}^3C_2}{4!}+\dfrac{{}^4C_2}{5!}+\ldotsa=1+3!2C2​​+4!3C2​​+5!4C2​​+… and b=1+1C0+1C1+2C21!+2C0+3C1+4C22!+3C0+4C1+5C23!+…b=1+\dfrac{{}^1C_0+{}^1C_1+{}^2C_2}{1!}+\dfrac{{}^2C_0+{}^3C_1+{}^4C_2}{2!}+\dfrac{{}^3C_0+{}^4C_1+{}^5C_2}{3!}+\ldotsb=1+1!1C0​+1C1​+2C2​​+2!2C0​+3C1​+4C2​​+3!3C0​+4C1​+5C2​​+…. Then 2ba2\dfrac{2b}{a^2}a22b​ is equal to ___

Correct answer: 8

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2024
Let SnS_nSn​ denote the sum of the first nnn terms of an arithmetic progression. If S10=390S_{10}=390S10​=390 and the ratio of the tenth and the fifth terms is 15:715:715:7, then S15−S5S_{15}-S_5S15​−S5​ is equal to:
  1. (A)800
  2. (B)890
  3. (C)790
  4. (D)690

Correct answer: (C)

Step-by-step solution →
Q82·MathematicsNumericalJEE Main 2024
Let 3,7,11,15,…,4033,7,11,15,\ldots,4033,7,11,15,…,403 and 2,5,8,11,…,4042,5,8,11,\ldots,4042,5,8,11,…,404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___

Correct answer: 6699

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Q83·MathematicsSingle correctJEE Main 2024
Let 3,a,b,c3,a,b,c3,a,b,c be in A.P. and 3,a−1,b+1,c+93,a-1,b+1,c+93,a−1,b+1,c+9 be in G.P. Then, the arithmetic mean of a,b,ca,b,ca,b,c is:
  1. (A)−4-4−4
  2. (B)−1-1−1
  3. (C)131313
  4. (D)111111

Correct answer: (D)

Step-by-step solution →
Q84·MathematicsNumericalJEE Main 2024
If three successive terms of a G.P. with common ratio rrr (r>1)(r>1)(r>1) are the lengths of the sides of a triangle and [r][r][r] denotes the greatest integer less than or equal to rrr, then 3[r]+[−r]3[r]+[-r]3[r]+[−r] is equal to __________.

Correct answer: 1

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2024
The sum of the series 11−3⋅12+14+21−3⋅22+24+31−3⋅32+34+⋯\dfrac{1}{1-3\cdot 1^2+1^4}+\dfrac{2}{1-3\cdot 2^2+2^4}+\dfrac{3}{1-3\cdot 3^2+3^4}+\cdots1−3⋅12+141​+1−3⋅22+242​+1−3⋅32+343​+⋯ up to 10 terms is
  1. (A)45109\dfrac{45}{109}10945​
  2. (B)−45109-\dfrac{45}{109}−10945​
  3. (C)55109\dfrac{55}{109}10955​
  4. (D)−55109-\dfrac{55}{109}−10955​

Correct answer: (D)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2024
Let 2nd2^{\text{nd}}2nd, 8th8^{\text{th}}8th and 44th44^{\text{th}}44th terms of a non-constant A.P. be respectively the 1st1^{\text{st}}1st, 2nd2^{\text{nd}}2nd and 3rd3^{\text{rd}}3rd terms of G.P. If the first term of A.P. is 1, then the sum of first 20 terms is equal to
  1. (A)980
  2. (B)960
  3. (C)990
  4. (D)970

Correct answer: (D)

Step-by-step solution →
Q87·MathematicsNumericalJEE Main 2024
Let SnS_nSn​ be the sum to nnn-terms of an arithmetic progression 3,7,11,…3,7,11,\ldots3,7,11,…. If 40<6n(n+1)∑k=1nSk<4240<\dfrac{6}{n(n+1)}\displaystyle\sum_{k=1}^{n}S_k<4240<n(n+1)6​k=1∑n​Sk​<42, then nnn equals ______.

Correct answer: 9

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2024
Let SnS_nSn​ denote the sum of first nnn terms of an arithmetic progression. If S20=790S_{20}=790S20​=790 and S10=145S_{10}=145S10​=145, then S15−S5S_{15}-S_5S15​−S5​ is:
  1. (A)395395395
  2. (B)390390390
  3. (C)405405405
  4. (D)410410410

Correct answer: (A)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2024
Let aaa and bbb be two distinct positive real numbers. Let 11th11^{\text{th}}11th term of a GP, whose first term is aaa and third term is bbb, is equal to pthp^{\text{th}}pth term of another GP, whose first term is aaa and fifth term is bbb. Then ppp is equal to:
  1. (A)202020
  2. (B)252525
  3. (C)212121
  4. (D)242424

Correct answer: (C)

Step-by-step solution →
Q90·MathematicsNumericalJEE Main 2024
Let α=12+42+82+132+192+262+…\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\ldotsα=12+42+82+132+192+262+… upto 101010 terms and β=∑n=110n4\beta=\displaystyle\sum_{n=1}^{10}n^4β=n=1∑10​n4. If 4α−β=55k+404\alpha-\beta=55k+404α−β=55k+40, then kkk is equal to ___

Correct answer: 353

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2024
If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P., then the common ratio of the G.P. is equal to
  1. (A)7
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2024
If each term of a geometric progression a1,a2,a3,…a_1,a_2,a_3,\ldotsa1​,a2​,a3​,… with a1=18a_1=\dfrac{1}{8}a1​=81​ and a2≠a1a_2\neq a_1a2​=a1​, is the arithmetic mean of the next two terms and Sn=a1+a2+…+anS_n=a_1+a_2+\ldots+a_nSn​=a1​+a2​+…+an​, then S20−S18S_{20}-S_{18}S20​−S18​ is equal to:
  1. (A)2152^{15}215
  2. (B)−218-2^{18}−218
  3. (C)2182^{18}218
  4. (D)−215-2^{15}−215

Correct answer: (D)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2024
If log⁡ea\log_e aloge​a, log⁡eb\log_e bloge​b, log⁡ec\log_e cloge​c are in an A.P. and log⁡ea−log⁡e2b\log_e a-\log_e 2bloge​a−loge​2b, log⁡e2b−log⁡e3c\log_e 2b-\log_e 3cloge​2b−loge​3c, log⁡e3c−log⁡ea\log_e 3c-\log_e aloge​3c−loge​a are also in an A.P., then a:b:ca:b:ca:b:c is equal to:
  1. (A)9:6:49:6:49:6:4
  2. (B)16:4:116:4:116:4:1
  3. (C)25:10:425:10:425:10:4
  4. (D)6:3:26:3:26:3:2

Correct answer: (A)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2024
In an A.P., the sixth term a6=2a_6=2a6​=2. If the a1a4a5a_1 a_4 a_5a1​a4​a5​ is the greatest, then the common difference of the A.P., is equal to
  1. (A)32\dfrac{3}{2}23​
  2. (B)85\dfrac{8}{5}58​
  3. (C)23\dfrac{2}{3}32​
  4. (D)58\dfrac{5}{8}85​

Correct answer: (B)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2024
If α,β\alpha,\betaα,β are the roots of the equation x2−x−1=0x^2-x-1=0x2−x−1=0 and Sn=2023αn+2024βnS_n=2023\alpha^n+2024\beta^nSn​=2023αn+2024βn, then
  1. (A)2S12=S11+S102S_{12}=S_{11}+S_{10}2S12​=S11​+S10​
  2. (B)S12=S11+S10S_{12}=S_{11}+S_{10}S12​=S11​+S10​
  3. (C)2S11=S12+S102S_{11}=S_{12}+S_{10}2S11​=S12​+S10​
  4. (D)S11=S10+S12S_{11}=S_{10}+S_{12}S11​=S10​+S12​

Correct answer: (B)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2024
The number of common terms in the progressions 4,9,14,19,…4,9,14,19,\ldots4,9,14,19,…, up to 25th25^{\text{th}}25th term and 3,6,9,12,…3,6,9,12,\ldots3,6,9,12,…, up to 37th37^{\text{th}}37th term is:
  1. (A)9
  2. (B)5
  3. (C)7
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q97·MathematicsNumericalJEE Main 2024
If 8=3+14(3+p)+142(3+2p)+143(3+3p)+…∞8=3+\dfrac{1}{4}(3+p)+\dfrac{1}{4^2}(3+2p)+\dfrac{1}{4^3}(3+3p)+\ldots\infty8=3+41​(3+p)+421​(3+2p)+431​(3+3p)+…∞, then the value of ppp is __________.

Correct answer: 9

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2024
The 20th20^{\text{th}}20th term from the end of the progression 20,1914,1812,1734,…,−1291420,19\dfrac14,18\dfrac12,17\dfrac34,\ldots,-129\dfrac1420,1941​,1821​,1743​,…,−12941​ is :
  1. (A)−118-118−118
  2. (B)−110-110−110
  3. (C)−115-115−115
  4. (D)−100-100−100

Correct answer: (C)

Step-by-step solution →
Q99·MathematicsNumericalJEE Advanced 2023
Let 75⋯5⏞r77\overbrace{5\cdots5}^{r}775⋯5r7 denote the (r + 2) digit number where the first and the last digits are 7 and the remaining r digits are 5. Consider the sum S=77+757+7557+…+75⋯5⏞987S = 77 + 757 + 7557 + \ldots + 7\overbrace{5\cdots5}^{98}7S=77+757+7557+…+75⋯5987. If S=75⋯5⏞997+mnS = \frac{7\overbrace{5\cdots5}^{99}7 + m}{n}S=n75⋯5997+m​ , where m and n are natural numbers less than 3000, then the value of m + n is

Correct answer: 1219

Step-by-step solution →
Q100·MathematicsNumericalJEE Main 2023
If the sum of the series (12−13)+(122−12⋅3+132)+(123−122⋅3+12⋅32−133)+(124−123⋅3+122⋅32−12⋅33+134)+⋯\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2}\right) + \left(\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3}\right) + \left(\frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4}\right) + \cdots(21​−31​)+(221​−2⋅31​+321​)+(231​−22⋅31​+2⋅321​−331​)+(241​−23⋅31​+22⋅321​−2⋅331​+341​)+⋯ is αβ\frac{\alpha}{\beta}βα​, where α\alphaα and β\betaβ are co-prime, then α+3β\alpha + 3\betaα+3β is equal to....

Correct answer: 7

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Q101·MathematicsSingle correctJEE Main 2023
Let A1A_1A1​ and A2A_2A2​ be two arithmetic means and G1,G2,G3G_1, G_2, G_3G1​,G2​,G3​ be three geometric means of two distinct positive numbers. The G14+G24+G34+G12G32G_1^4+G_2^4+G_3^4+G_1^2 G_3^2G14​+G24​+G34​+G12​G32​ is equal to
  1. (A)2(A1+A2)G1G32(A_1+A_2)G_1 G_32(A1​+A2​)G1​G3​
  2. (B)(A1+A2)2G1G3(A_1+A_2)^2 G_1 G_3(A1​+A2​)2G1​G3​
  3. (C)(A1+A2)G12G32(A_1+A_2)G_1^2 G_3^2(A1​+A2​)G12​G32​
  4. (D)2(A1+A2)G12G322(A_1+A_2)G_1^2 G_3^22(A1​+A2​)G12​G32​

Correct answer: (B)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2023
Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be a G.P. of increasing positive numbers. Let the sum of its 6th6^{th}6th and 8th8^{th}8th terms be 2 and the product of its 3rd3^{rd}3rd and 5th5^{th}5th terms be 19\dfrac{1}{9}91​. Then 6(a2+a4)(a4+a6)6(a_2 + a_4)(a_4 + a_6)6(a2​+a4​)(a4​+a6​) is equal to
  1. (A)222\sqrt{2}22​
  2. (B)222
  3. (C)333\sqrt{3}33​
  4. (D)333

Correct answer: (D)

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
Let [α][\alpha][α] denote the greatest integer ≤α\le \alpha≤α. Then [1]+[2]+[3]+…+[120]\left[\sqrt{1}\right] + \left[\sqrt{2}\right] + \left[\sqrt{3}\right] + \ldots + \left[\sqrt{120}\right][1​]+[2​]+[3​]+…+[120​] is equal to

Correct answer: 825

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
Among: (S1): lim⁡n→∞1n2(2+4+6+⋯+2n)=1\displaystyle\lim_{n\to\infty}\dfrac{1}{n^{2}}(2+4+6+\dots+2n)=1n→∞lim​n21​(2+4+6+⋯+2n)=1 (S2): lim⁡n→∞1n16(115+215+315+⋯+n15)=116\displaystyle\lim_{n\to\infty}\dfrac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\dots+n^{15})=\dfrac{1}{16}n→∞lim​n161​(115+215+315+⋯+n15)=161​
  1. (A)Both (S1) and (S2) are false
  2. (B)Both (S1) and (S2) are true
  3. (C)Only (S2) is true
  4. (D)Only (S1) is true

Correct answer: (A)

Step-by-step solution →
Q105·MathematicsNumericalJEE Main 2023
The sum to 202020 terms of the series 2⋅22−32+2⋅42−52+2⋅62−…2\cdot 2^{2}-3^{2}+2\cdot 4^{2}-5^{2}+2\cdot 6^{2}-\dots2⋅22−32+2⋅42−52+2⋅62−… is equal to _____.

Correct answer: 1310

Step-by-step solution →
Q106·MathematicsNumericalJEE Main 2023
Let f(x)=∑k=110kxkf(x) = \sum\limits_{k=1}^{10} k x^kf(x)=k=1∑10​kxk, x∈Rx \in \mathbb{R}x∈R. If 2f(2)+f′(2)=119(2)k+12f(2) + f'(2) = 119(2)^k + 12f(2)+f′(2)=119(2)k+1 then kkk is equal to

Correct answer: 10

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2023
Let s1,s2,s3,…,s10s_{1},s_{2},s_{3},\dots,s_{10}s1​,s2​,s3​,…,s10​ respectively be the sum to 121212 terms of 101010 A.P.'s whose first terms are 1,2,3,…,101,2,3,\dots,101,2,3,…,10 and the common differences are 1,3,5,…,191,3,5,\dots,191,3,5,…,19 respectively. Then ∑i=110si\displaystyle\sum_{i=1}^{10}s_{i}i=1∑10​si​ is equal to:
  1. (A)738073807380
  2. (B)722072207220
  3. (C)736073607360
  4. (D)726072607260

Correct answer: (D)

Step-by-step solution →
Q108·MathematicsNumericalJEE Main 2023
Let the positive numbers a1,a2,a3,a4a_1,a_2,a_3,a_4a1​,a2​,a3​,a4​ and a5a_5a5​ be in a G.P. Let their mean and variance be 3110\dfrac{31}{10}1031​ and mn\dfrac{m}{n}nm​ respectively, where m and n are co-prime. If the mean of their reciprocals is 3140\dfrac{31}{40}4031​ and a3+a4+a5=14a_3+a_4+a_5=14a3​+a4​+a5​=14, then m+nm+nm+n is equal to _________.

Correct answer: 211

Step-by-step solution →
Q109·MathematicsSingle correctJEE Main 2023
Let ⟨an⟩\langle a_n\rangle⟨an​⟩ be a sequence such that a1+a2+⋯+an=n2+3n(n+1)(n+2)a_1+a_2+\cdots+a_n=\dfrac{n^2+3n}{(n+1)(n+2)}a1​+a2​+⋯+an​=(n+1)(n+2)n2+3n​. If 28∑k=1101ak=p1p2p3⋯pm28\sum_{k=1}^{10}\dfrac{1}{a_k}=p_1 p_2 p_3\cdots p_m28∑k=110​ak​1​=p1​p2​p3​⋯pm​, where p1,p2,…,pmp_1,p_2,\ldots,p_mp1​,p2​,…,pm​ are the first m prime numbers, then m is equal to
  1. (A)7
  2. (B)6
  3. (C)5
  4. (D)8

Correct answer: (B)

Step-by-step solution →
Q110·MathematicsNumericalJEE Main 2023
For k∈Nk\in\mathbb{N}k∈N, if the sum of the series 1+4k+8k2+13k3+19k4+⋯1+\dfrac{4}{k}+\dfrac{8}{k^2}+\dfrac{13}{k^3}+\dfrac{19}{k^4}+\cdots1+k4​+k28​+k313​+k419​+⋯ is 101010, then the value of kkk is

Correct answer: 2

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2023
Let x1,x2,…,x100x_1, x_2, \ldots, x_{100}x1​,x2​,…,x100​ be in an arithmetic progression, with x1=2x_1 = 2x1​=2 and their mean equal to 200200200. If yi=i(xi−i)y_i = i(x_i - i)yi​=i(xi​−i), 1≤i≤1001 \le i \le 1001≤i≤100, then the mean of y1,y2,…,y100y_1, y_2, \ldots, y_{100}y1​,y2​,…,y100​ is
  1. (A)10101.5010101.5010101.50
  2. (B)10051.5010051.5010051.50
  3. (C)10049.5010049.5010049.50
  4. (D)101001010010100

Correct answer: (C)

Step-by-step solution →
Q112·MathematicsNumericalJEE Main 2023
Let S=109+1085+10752+…+25107+15108S = 109 + \frac{108}{5} + \frac{107}{5^2} + \ldots + \frac{2}{5^{107}} + \frac{1}{5^{108}}S=109+5108​+52107​+…+51072​+51081​. Then the value of (16S−(25)−54)(16S - (25)^{-54})(16S−(25)−54) is equal to _______ .

Correct answer: 2175

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2023
If Sn=4+11+21+34+50+…S_n=4+11+21+34+50+\ldotsSn​=4+11+21+34+50+… to nnn terms, then 160(S29−S6)\frac{1}{60}\left(S_{29}-S_6\right)601​(S29​−S6​) is equal to:
  1. (A)226226226
  2. (B)220220220
  3. (C)223223223
  4. (D)227227227

Correct answer: (C)

Step-by-step solution →
Q114·MathematicsNumericalJEE Main 2023
Suppose a1,a2,2,a3,a4a_1,a_2,2,a_3,a_4a1​,a2​,2,a3​,a4​ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 222 and the sum of all 555 terms of the arithmetico-geometric progression is 492\frac{49}{2}249​, then a4a_4a4​ is equal to _______ .

Correct answer: 16

Step-by-step solution →
Q115·MathematicsNumericalJEE Main 2023
The sum of all those terms, of the arithmetic progression 3,8,13,…,3733,8,13,\ldots,3733,8,13,…,373, which are not divisible by 3, is equal to _________.

Correct answer: 9525

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2023
Let the first term aaa and the common ratio rrr of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
  1. (A)231231231
  2. (B)210210210
  3. (C)220220220
  4. (D)241241241

Correct answer: (A)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2023
Let ana_nan​ be the nthn^{th}nth term of the series 5+8+14+23+35+50+…5+8+14+23+35+50+\ldots5+8+14+23+35+50+… and Sn=∑k=1nakS_n=\sum_{k=1}^{n}a_kSn​=∑k=1n​ak​. Then S30−a40S_{30}-a_{40}S30​−a40​ is equal to
  1. (A)11310
  2. (B)11280
  3. (C)11290
  4. (D)11260

Correct answer: (C)

Step-by-step solution →
Q118·MathematicsNumericalJEE Main 2023
If ana_{n}an​ is the greatest term in the sequence an=n3n4+147a_{n}=\frac{n^{3}}{n^{4}+147}an​=n4+147n3​, n=1,2,3,…n=1,2,3,\ldotsn=1,2,3,…, then α\alphaα is equal to

Correct answer: 5

Step-by-step solution →
Q119·MathematicsNumericalJEE Main 2023
Let 0<z<y<x0<z<y<x0<z<y<x be three real numbers such that 1x,1y,1z\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}x1​,y1​,z1​ are in an arithmetic progression and x,2y,zx,\sqrt{2}y,zx,2​y,z are in a geometric progression. If xy+yz+zx=32xyzxy+yz+zx=\dfrac{3}{\sqrt{2}}xyzxy+yz+zx=2​3​xyz, then 3(x+y+z)23(x+y+z)^{2}3(x+y+z)2 is equal to

Correct answer: 150

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2023
Let Sk=1+2+⋯+KKS_{k}=\frac{1+2+\dots+K}{K}Sk​=K1+2+⋯+K​ and ∑j=1nSj2=nA(Bn2+Cn+D)\sum_{j=1}^{n} S_{j}^{2}=\frac{n}{A}(Bn^{2}+Cn+D)∑j=1n​Sj2​=An​(Bn2+Cn+D), where A,B,C,D∈NA,B,C,D\in NA,B,C,D∈N and AAA has least value. Then
  1. (A)A+BA+BA+B is divisible by DDD
  2. (B)A+B=5(D−C)A+B=5(D-C)A+B=5(D−C)
  3. (C)A+C+DA+C+DA+C+D is not divisible by BBB
  4. (D)A+B+C+DA+B+C+DA+B+C+D is divisible by 5

Correct answer: (A)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2023
The sum of the first 202020 terms of the series 5+11+19+29+41+…5+11+19+29+41+\dots5+11+19+29+41+… is:
  1. (A)3450
  2. (B)3250
  3. (C)3420
  4. (D)3520

Correct answer: (D)

Step-by-step solution →
Q122·MathematicsNumericalJEE Main 2023
If (20)19+2(21)(20)18+3(21)2(20)17+⋯+20(21)19=k(20)18(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17}+\cdots+20(21)^{19}=k(20)^{18}(20)19+2(21)(20)18+3(21)2(20)17+⋯+20(21)19=k(20)18 then kkk is equal to

Correct answer: 400

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2023
Let a1,a2,…,ana_1,a_2,\dots,a_na1​,a2​,…,an​ be nnn positive consecutive terms of an arithmetic progression. If d>0d>0d>0 is its common difference, then lim⁡n→∞dn(1a1+a2+1a2+a3+⋯+1an−1+an)\displaystyle\lim_{n\to\infty}\sqrt{\dfrac{d}{n}}\left(\dfrac{1}{\sqrt{a_1}+\sqrt{a_2}}+\dfrac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)n→∞lim​nd​​(a1​​+a2​​1​+a2​​+a3​​1​+⋯+an−1​​+an​​1​) is:
  1. (A)d\sqrt dd​
  2. (B)000
  3. (C)1d\dfrac{1}{\sqrt d}d​1​
  4. (D)∞\infty∞

Correct answer: (A)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2023
If gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1 and 12−22+32−42+⋯+(2021)2−(2022)2+(2023)2=1012 m2n1^2-2^2+3^2-4^2+\cdots+(2021)^2-(2022)^2+(2023)^2=1012\,m^2n12−22+32−42+⋯+(2021)2−(2022)2+(2023)2=1012m2n then m2−n2m^2-n^2m2−n2 is equal to
  1. (A)240
  2. (B)220
  3. (C)210
  4. (D)180

Correct answer: (D)

Step-by-step solution →
Q125·MathematicsNumericalJEE Main 2023
The sum of the common terms of the following three arithmetic progressions: 3,7,11,15,…,3993,7,11,15,\ldots,3993,7,11,15,…,399; 2,5,8,11,…,3592,5,8,11,\ldots,3592,5,8,11,…,359 and 2,7,12,17,…,1972,7,12,17,\ldots,1972,7,12,17,…,197 is equal to

Correct answer: 321

Step-by-step solution →
Q126·MathematicsNumericalJEE Main 2023
Let a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_na1​,a2​,a3​,…,an​ be in A.P. If the sum of its first four terms is 505050 and the sum of its last four terms is 170170170, then the product of its middle two terms is _______ .

Correct answer: 754

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2023
The sum to 101010 terms of the series 11+12+14+21+22+24+31+32+34+⋯\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\cdots1+12+141​+1+22+242​+1+32+343​+⋯ is
  1. (A)55111\frac{55}{111}11155​
  2. (B)56111\frac{56}{111}11156​
  3. (C)58111\frac{58}{111}11158​
  4. (D)59111\frac{59}{111}11159​

Correct answer: (A)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2023
The sum ∑n=1∞2n2+3n+4(2n)!\sum_{n=1}^{\infty}\dfrac{2n^2+3n+4}{(2n)!}∑n=1∞​(2n)!2n2+3n+4​ is equal to:
  1. (A)13e4+54e\dfrac{13e}{4}+\dfrac{5}{4e}413e​+4e5​
  2. (B)11e2+72e−4\dfrac{11e}{2}+\dfrac{7}{2e}-4211e​+2e7​−4
  3. (C)11e2+72e\dfrac{11e}{2}+\dfrac{7}{2e}211e​+2e7​
  4. (D)13e4+54e−4\dfrac{13e}{4}+\dfrac{5}{4e}-4413e​+4e5​−4

Correct answer: (D)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2023
Let a1,a2,…,ana_1, a_2, \ldots, a_na1​,a2​,…,an​ be in A.P. If a5=2a7a_5 = 2a_7a5​=2a7​ and a11=18a_{11} = 18a11​=18, then 12(1a10+a11+1a11+a12+⋯+1a17+a18)12\left(\frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}}\right)12(a10​​+a11​​1​+a11​​+a12​​1​+⋯+a17​​+a18​​1​) is equal to

Correct answer: 8

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2023
Let a1,a2,a3,…a_1,a_2,a_3,\ldotsa1​,a2​,a3​,… be an A.P. If a7=3a_7=3a7​=3, the product a1a4a_1a_4a1​a4​ is minimum and the sum of its first nnn terms is zero, then n!−4an(n+2)n!-4a_{n(n+2)}n!−4an(n+2)​ is equal to:
  1. (A)999
  2. (B)334\dfrac{33}{4}433​
  3. (C)3814\dfrac{381}{4}4381​
  4. (D)242424

Correct answer: (D)

Step-by-step solution →
Q131·MathematicsNumericalJEE Main 2023
The sum 12−2⋅32+3⋅52−4⋅72+5⋅92−⋯+15⋅2921^2-2\cdot3^2+3\cdot5^2-4\cdot7^2+5\cdot9^2-\cdots+15\cdot29^212−2⋅32+3⋅52−4⋅72+5⋅92−⋯+15⋅292 is

Correct answer: 6952

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2023
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is
  1. (A)777
  2. (B)333
  3. (C)92\frac{9}{2}29​
  4. (D)141414

Correct answer: (A)

Step-by-step solution →
Q133·MathematicsSingle correctJEE Main 2023
Let α∈(0,1)\alpha \in (0,1)α∈(0,1) and β=log⁡e(1−α)\beta = \log_e(1 - \alpha)β=loge​(1−α). Let Pn(x)=x+x22+x33+⋯+xnn, x∈(0,1)P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n},\ x \in (0,1)Pn​(x)=x+2x2​+3x3​+⋯+nxn​, x∈(0,1). Then the integral ∫0αt501−t dt\int_0^{\alpha} \frac{t^{50}}{1 - t}\, dt∫0α​1−tt50​dt is equal to
  1. (A)β+P50(α)\beta + P_{50}(\alpha)β+P50​(α)
  2. (B)P50(α)−βP_{50}(\alpha) - \betaP50​(α)−β
  3. (C)β−P50(α)\beta - P_{50}(\alpha)β−P50​(α)
  4. (D)−(β+P50(α))-(\beta + P_{50}(\alpha))−(β+P50​(α))

Correct answer: (D)

Step-by-step solution →
Q134·MathematicsNumericalJEE Main 2023
Let ∑n=0∞n3((2n)!)+(2n−1)(n!)(n!)((2n)!)=ae+be+c\sum_{n=0}^{\infty}\dfrac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}=ae+\dfrac{b}{e}+c∑n=0∞​(n!)((2n)!)n3((2n)!)+(2n−1)(n!)​=ae+eb​+c, where a,b,c∈Za,b,c\in\mathbb{Z}a,b,c∈Z and e=∑n=0∞1n!e=\sum_{n=0}^{\infty}\dfrac{1}{n!}e=∑n=0∞​n!1​. Then a2−b+ca^2-b+ca2−b+c is equal to

Correct answer: 26

Step-by-step solution →
Q135·MathematicsNumericalJEE Main 2023
The 8th8^{th}8th common term of the series S1=3+7+11+15+19+⋯S_1=3 + 7 + 11 + 15 + 19 + \cdotsS1​=3+7+11+15+19+⋯ and S2=1+6+11+16+21+⋯S_2=1 + 6 + 11 + 16 + 21 + \cdotsS2​=1+6+11+16+21+⋯ is _______.

Correct answer: 151

Step-by-step solution →
Q136·MathematicsSingle correctJEE Main 2023
Let a,b,c>1a, b, c > 1a,b,c>1; a3,b3a^3, b^3a3,b3 and c3c^3c3 be in A.P., and log⁡ab,log⁡ca\log_a b, \log_c aloga​b,logc​a and log⁡bc\log_b clogb​c be in G.P. If the sum of the first 20 terms of an A.P., whose first term is a+4b+c3\dfrac{a + 4b + c}{3}3a+4b+c​ and the common difference is a−8b+c10\dfrac{a - 8b + c}{10}10a−8b+c​, is −444-444−444, then abcabcabc is equal to:
  1. (A)1258\dfrac{125}{8}8125​
  2. (B)216216216
  3. (C)343343343
  4. (D)3438\dfrac{343}{8}8343​

Correct answer: (B)

Step-by-step solution →
Q137·MathematicsSingle correctJEE Main 2023
If an=−24n2−16n+15a_n=\dfrac{-2}{4n^2-16n+15}an​=4n2−16n+15−2​, then a1+a2+⋯+a25a_1+a_2+\cdots+a_{25}a1​+a2​+⋯+a25​ is equal to:
  1. (A)52147\dfrac{52}{147}14752​
  2. (B)49138\dfrac{49}{138}13849​
  3. (C)50141\dfrac{50}{141}14150​
  4. (D)51144\dfrac{51}{144}14451​

Correct answer: (C)

Step-by-step solution →
Q138·MathematicsSingle correctJEE Main 2023
Suppose f:R→(0,∞)f:\mathbb{R}\to(0,\infty)f:R→(0,∞) be a differentiable function such that 5f(x+y)=f(x)⋅f(y)5f(x+y)=f(x)\cdot f(y)5f(x+y)=f(x)⋅f(y), ∀x,y∈R\forall x,y\in\mathbb{R}∀x,y∈R. If f(3)=320f(3)=320f(3)=320, then ∑n=05f(n)\sum_{n=0}^{5}f(n)∑n=05​f(n) is equal to:
  1. (A)687568756875
  2. (B)652565256525
  3. (C)682568256825
  4. (D)657565756575

Correct answer: (C)

Step-by-step solution →
Q139·MathematicsNumericalJEE Main 2023
Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then a1a9+a2a4a9+a5+a7a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7a1​a9​+a2​a4​a9​+a5​+a7​ is equal to ________ .

Correct answer: 60

Step-by-step solution →
Q140·MathematicsNumericalJEE Main 2023
Let a1=b1=1a_1=b_1=1a1​=b1​=1 and an=an−1+(n−1), bn=bn−1+an−1, ∀n≥2a_n=a_{n-1}+(n-1),\,b_n=b_{n-1}+a_{n-1},\,\forall n\ge 2an​=an−1​+(n−1),bn​=bn−1​+an−1​,∀n≥2. If S=∑n=110bn2nS=\sum_{n=1}^{10}\dfrac{b_n}{2^n}S=∑n=110​2nbn​​ and T=∑n=18n2n−1T=\sum_{n=1}^{8}\dfrac{n}{2^{n-1}}T=∑n=18​2n−1n​, then 27(2S−T)2^{7}(2S-T)27(2S−T) is equal to _____.

Correct answer: 461

Step-by-step solution →
Q141·MathematicsNumericalJEE Main 2023
Let {ak}\{a_k\}{ak​} and {bk}, k∈N\{b_k\},\,k\in\mathbb{N}{bk​},k∈N, be two G.P.s with common ratios r1r_1r1​ and r2r_2r2​ respectively such that a1=b1=4a_1=b_1=4a1​=b1​=4 and r1<r2r_1<r_2r1​<r2​. Let ck=ak+bk, k∈Nc_k=a_k+b_k,\,k\in\mathbb{N}ck​=ak​+bk​,k∈N. If c2=5c_2=5c2​=5 and c3=134c_3=\dfrac{13}{4}c3​=413​ then ∑k=1∞ck−(12a6+8b4)\sum_{k=1}^{\infty}c_k-(12a_6+8b_4)∑k=1∞​ck​−(12a6​+8b4​) is equal to _____.

Correct answer: 9

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2023
Suppose fff is a function satisfying f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)f(x+y)=f(x)+f(y) for all x,y∈Nx, y \in \mathbb{N}x,y∈N and f(1)=15f(1) = \dfrac{1}{5}f(1)=51​. If ∑n=1mf(n)n(n+1)(n+2)=112\sum_{n=1}^{m} \dfrac{f(n)}{n(n + 1)(n + 2)} = \dfrac{1}{12}∑n=1m​n(n+1)(n+2)f(n)​=121​, then mmm is equal to ________ .

Correct answer: 10

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2023
Consider a function f:N→Rf:\mathbb{N}\to\mathbb{R}f:N→R, satisfying f(1)+2f(2)+3f(3)+⋯+xf(x)=x(x+1)f(x); x≥2f(1)+2f(2)+3f(3)+\dots+xf(x)=x(x+1)f(x);\,x\ge 2f(1)+2f(2)+3f(3)+⋯+xf(x)=x(x+1)f(x);x≥2 with f(1)=1f(1)=1f(1)=1. Then 1f(2022)+1f(2028)\dfrac{1}{f(2022)}+\dfrac{1}{f(2028)}f(2022)1​+f(2028)1​ is equal to:
  1. (A)810081008100
  2. (B)840084008400
  3. (C)800080008000
  4. (D)820082008200

Correct answer: (A)

Step-by-step solution →
Q144·MathematicsNumericalJEE Main 2023
Let A1,A2,A3A_1,A_2,A_3A1​,A2​,A3​ be three A.P. with the same common difference ddd and having their first terms as A,A+1,A+2A,A+1,A+2A,A+1,A+2, respectively. Let a,b,ca,b,ca,b,c be the 7th,9th,17th7^{th},9^{th},17^{th}7th,9th,17th terms of A1,A2,A3A_1,A_2,A_3A1​,A2​,A3​, respectively, such that ∣a712b171c171∣+70=0\begin{vmatrix}a & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1\end{vmatrix}+70=0​a2bc​71717​111​​+70=0. If a=29a=29a=29, then the sum of the first 202020 terms of an AP whose first term is c−a−bc-a-bc−a−b and common difference is d12\dfrac{d}{12}12d​, is equal to _______.

Correct answer: 495

Step-by-step solution →
Q145·MathematicsNumericalJEE Main 2023
For two positive numbers a,ba,ba,b such that a,ba,ba,b and 118\dfrac{1}{18}181​ are in a geometric progression, while 1a,10\dfrac{1}{a},10a1​,10 and 1b\dfrac{1}{b}b1​ are in an arithmetic progression, then 16a+b16a+b16a+b is equal to

Correct answer: 3

Step-by-step solution →
Q146·MathematicsNumericalJEE Main 2023
If 13+23+33+⋯ up to n terms1⋅3+2⋅5+3⋅7+⋯ up to n terms=95\dfrac{1^3+2^3+3^3+\cdots\text{ up to }n\text{ terms}}{1\cdot3+2\cdot5+3\cdot7+\cdots\text{ up to }n\text{ terms}}=\dfrac{9}{5}1⋅3+2⋅5+3⋅7+⋯ up to n terms13+23+33+⋯ up to n terms​=59​, then the value of nnn is

Correct answer: 5

Step-by-step solution →
Q147·MathematicsNumericalJEE Main 2023
The 4th4^{\text{th}}4th term of GP is 500500500 and its common ratio is 1m\dfrac{1}{m}m1​, m∈Nm\in\mathbb{N}m∈N. Let SnS_nSn​ denote the sum of the first nnn terms of this GP. If S6>S5+1S_6>S_5+1S6​>S5​+1 and S7<S6+12S_7<S_6+\tfrac{1}{2}S7​<S6​+21​, then the number of possible values of mmm is _______ .

Correct answer: 12

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2023
For three positive integers p,q,rp,q,rp,q,r, xpq2=yqr=zp2rx^{pq^2}=y^{qr}=z^{p^2 r}xpq2=yqr=zp2r and r=pq+1r=pq+1r=pq+1 such that 3, 3log⁡yx, 3log⁡zy, 7log⁡xz3,\ 3\log_y x,\ 3\log_z y,\ 7\log_x z3, 3logy​x, 3logz​y, 7logx​z are in A.P. with common difference 12\tfrac{1}{2}21​. Then r−p−qr-p-qr−p−q is equal to
  1. (A)−6-6−6
  2. (B)121212
  3. (C)666
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2023
Let f(x)f(x)f(x) be a function such that f(x+y)=f(x)⋅f(y)f(x+y)=f(x)\cdot f(y)f(x+y)=f(x)⋅f(y) for all x,y∈Nx,y\in\mathbb{N}x,y∈N. If f(1)=3f(1)=3f(1)=3 and ∑k=1nf(k)=3279\sum_{k=1}^{n}f(k)=3279∑k=1n​f(k)=3279, then the value of nnn is
  1. (A)999
  2. (B)666
  3. (C)888
  4. (D)777

Correct answer: (D)

Step-by-step solution →
Q150·MathematicsMultiple correctJEE Advanced 2022
Let a1,a2,a3,.....a_{1}, a_{2}, a_{3}, .....a1​,a2​,a3​,..... be an arithmetic progression with a1=7a_{1} = 7a1​=7 and common difference 8. Let T1,T2,T3,.....T_{1}, T_{2}, T_{3}, .....T1​,T2​,T3​,..... be such that T1=3T_{1} = 3T1​=3 and Tn+1−Tn=anT_{n+1} - T_{n} = a_{n}Tn+1​−Tn​=an​ for n≥1n \geq 1n≥1. Then, which of the following is/are TRUE?
  1. (A)T20=1604T_{20} = 1604T20​=1604
  2. (B)∑k=120Tk=10510\sum_{k=1}^{20} T_{k} = 10510∑k=120​Tk​=10510
  3. (C)T30=3454T_{30} = 3454T30​=3454
  4. (D)∑k=130Tk=35610\sum_{k=1}^{30} T_{k} = 35610∑k=130​Tk​=35610

Correct answer: (B), (C)

Step-by-step solution →
Q151·MathematicsNumericalJEE Advanced 2022
Let l1,l2,.....,l100l_{1}, l_{2}, ....., l_{100}l1​,l2​,.....,l100​ be consecutive terms of an arithmetic with common difference d1d_{1}d1​, and let w1,w2,.....,w100w_{1}, w_{2}, ....., w_{100}w1​,w2​,.....,w100​ be consecutive terms of another arithmetic progression with common difference d2d_{2}d2​, where d1d2=10d_{1}d_{2} = 10d1​d2​=10. For each iii = 1, 2, ....., 100, let RiR_{i}Ri​ be a rectangle with length lil_{i}li​, width wiw_{i}wi​ and area AiA_{i}Ai​. If A51−A50=1000A_{51} - A_{50} = 1000A51​−A50​=1000, then the value of A100−A90A_{100} - A_{90}A100​−A90​ is __________.

Correct answer: 18900

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2022
If 1(20−a) (40−a)+1(40−a) (60−a)+……+1(180−a) (200−a)=1256\frac{1}{(20 - a)\,(40 - a)} + \frac{1}{(40 - a)\,(60 - a)} + \ldots\ldots+ \frac{1}{(180 - a)\,(200 - a)} = \frac{1}{256}(20−a)(40−a)1​+(40−a)(60−a)1​+……+(180−a)(200−a)1​=2561​, then the maximum value of a is :
  1. (A)198
  2. (B)202
  3. (C)212
  4. (D)218

Correct answer: (C)

Step-by-step solution →
Q153·MathematicsSingle correctJEE Main 2022
Let {an}n=0∞\{a_n\}_{n=0}^{\infty}{an​}n=0∞​ be a sequence such that a0=a1=0a_0 = a_1 = 0a0​=a1​=0 and an+2=3an+1−2an+1a_{n+2} = 3a_{n+1} - 2a_n + 1an+2​=3an+1​−2an​+1, ∀ n≥0\forall\ n \geq 0∀ n≥0. Then a25 a23−2 a25 a22−2 a23 a24+4 a22 a24a_{25}\, a_{23} - 2\, a_{25}\, a_{22} - 2\, a_{23}\, a_{24} + 4\, a_{22}\, a_{24}a25​a23​−2a25​a22​−2a23​a24​+4a22​a24​ is equal to:
  1. (A)483483483
  2. (B)528528528
  3. (C)575575575
  4. (D)624624624

Correct answer: (B)

Step-by-step solution →
Q154·MathematicsNumericalJEE Main 2022
Let a1, a2, a3,…a_{1},\,a_{2},\,a_{3},\ldotsa1​,a2​,a3​,… be an A.P. If ∑r=1∞ar2r=4\sum_{r=1}^{\infty}\frac{a_{r}}{2^{r}} = 4∑r=1∞​2rar​​=4, then 4a24a_{2}4a2​ is equal to __________.

Correct answer: 16

Step-by-step solution →
Q155·MathematicsNumericalJEE Main 2022
If 12×3×4+13×4×5+14×5×6+…+1100×101×102=k101\frac{1}{2 \times 3 \times 4} + \frac{1}{3 \times 4 \times 5} + \frac{1}{4 \times 5 \times 6} + \ldots + \frac{1}{100 \times 101 \times 102} = \frac{k}{101}2×3×41​+3×4×51​+4×5×61​+…+100×101×1021​=101k​, then 34 k is equal to _____.

Correct answer: 286

Step-by-step solution →
Q156·MathematicsSingle correctJEE Main 2022
∑r=120(r2+1)(r!)\sum_{r=1}^{20} \left( r^2 + 1 \right)\left( r! \right)∑r=120​(r2+1)(r!) is equal to:
  1. (A)22!−21!22! - 21!22!−21!
  2. (B)22!−2 (21!)22! - 2\,(21!)22!−2(21!)
  3. (C)21!−2 (20!)21! - 2\,(20!)21!−2(20!)
  4. (D)21!−20!21! - 20!21!−20!

Correct answer: (B)

Step-by-step solution →
Q157·MathematicsNumericalJEE Main 2022
If 6312+10311+20310+4039+…+102403=2n⋅m\frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^{9}}+\ldots+\frac{10240}{3}=2^{n}\cdot m3126​+31110​+31020​+3940​+…+310240​=2n⋅m, where m is odd, then m.nm.nm.n is equal to _____

Correct answer: 12

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2022
Consider the sequence a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​, …… such that a1=1a_{1} = 1a1​=1, a2=2a_{2} = 2a2​=2 and an+2=2an+1+ana_{n+2} = \frac{2}{a_{n+1}} + a_{n}an+2​=an+1​2​+an​ for n = 1, 2, 3, … . If (a1+1a2a3)⋅(a2+1a3a4)⋅(a3+1a4a5)⋯(a30+1a31a32)=2α(61C31)\left( \frac{a_{1} + \frac{1}{a_{2}}}{a_{3}} \right) \cdot \left( \frac{a_{2} + \frac{1}{a_{3}}}{a_{4}} \right) \cdot \left( \frac{a_{3} + \frac{1}{a_{4}}}{a_{5}} \right) \cdots \left( \frac{a_{30} + \frac{1}{a_{31}}}{a_{32}} \right) = 2^{\alpha} \left( {}^{61}C_{31} \right)(a3​a1​+a2​1​​)⋅(a4​a2​+a3​1​​)⋅(a5​a3​+a4​1​​)⋯(a32​a30​+a31​1​​)=2α(61C31​), then α\alphaα is equal to :
  1. (A)−30
  2. (B)−31
  3. (C)−60
  4. (D)−61

Correct answer: (C)

Step-by-step solution →
Q159·MathematicsNumericalJEE Main 2022
23−131×7+43−33+23−132×11+63−53+43−33+23−133×15+.....+303−293+283−273+...+23−1315×63\frac{2^{3}-1^{3}}{1 \times 7} + \frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11} + \frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15} + ..... + \frac{30^{3}-29^{3}+28^{3}-27^{3}+...+2^{3}-1^{3}}{15 \times 63}1×723−13​+2×1143−33+23−13​+3×1563−53+43−33+23−13​+.....+15×63303−293+283−273+...+23−13​ is equal to _____.

Correct answer: 120

Step-by-step solution →
Q160·MathematicsSingle correctJEE Main 2022
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 9825\frac{98}{25}2598​. Then the sum of the first 21 terms of an AP, whose first term is 10ar10ar10ar, nthn^{th}nth term is ana_{n}an​ and the common difference is 10ar210ar^{2}10ar2, is equal to :
  1. (A)21 a1121\,a_{11}21a11​
  2. (B)22 a1122\,a_{11}22a11​
  3. (C)15 a1615\,a_{16}15a16​
  4. (D)14 a1614\,a_{16}14a16​

Correct answer: (A)

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2022
Suppose a1,a2a_1, a_2a1​,a2​, ...., ana_nan​,... be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of the sum of first nine terms of the progression is 5 : 17 and 110<a15<120110 < a_{15} < 120110<a15​<120, then the sum of the first ten terms of the progression is equal to -
  1. (A)290
  2. (B)380
  3. (C)460
  4. (D)510

Correct answer: (B)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2022
Consider two G.Ps. 2, 22^{2}2, 23^{3}3, ... and 4, 42^{2}2, 43^{3}3,.... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is (2)2258(2)^{\frac{225}{8}}(2)8225​ , then ∑k=1nk(n−k)\sum\limits_{k=1}^{n} k(n-k)k=1∑n​k(n−k) is equal to :
  1. (A)560
  2. (B)1540
  3. (C)1330
  4. (D)2600

Correct answer: (C)

Step-by-step solution →
Q163·MathematicsNumericalJEE Main 2022
If ∑k=110kk4+k2+1=mn\sum_{k=1}^{10}\frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}∑k=110​k4+k2+1k​=nm​, where m and n are co-prime, then m+nm+nm+n is equal to

Correct answer: 166

Step-by-step solution →
Q164·MathematicsNumericalJEE Main 2022
The series of positive multiples of 3 is divided into sets : {3}, {6, 9,12}, {15, 18, 21, 24, 27},... Then the sum of the elements in the 11th^{th}th set is equal to__________,

Correct answer: 6993

Step-by-step solution →
Q165·MathematicsNumericalJEE Main 2022
Different A.P.'s are constructed with the first term 100, the last term 199, And integral common differences. The sum of the common differences of all such, A.P's having at least 3 terms and at most 33 terms is.

Correct answer: 53

Step-by-step solution →
Q166·MathematicsNumericalJEE Main 2022
Let a1=b1=1a_{1} = b_{1} = 1a1​=b1​=1, an=an−1+2a_{n} = a_{n-1} + 2an​=an−1​+2 and bn=an+bn−1b_{n} = a_{n} + b_{n-1}bn​=an​+bn−1​ for every natural number n≥2n \ge 2n≥2. Then ∑n=115an⋅bn\sum_{n=1}^{15} a_{n} \cdot b_{n}∑n=115​an​⋅bn​ is equal to ________.

Correct answer: 27560

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2022
The sum of the infinite series 1+56+1262+2263+3564+5165+7066+....1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \frac{51}{6^{5}} + \frac{70}{6^{6}} + ....1+65​+6212​+6322​+6435​+6551​+6670​+.... is equal to:
  1. (A)425216\frac{425}{216}216425​
  2. (B)429216\frac{429}{216}216429​
  3. (C)288125\frac{288}{125}125288​
  4. (D)280125\frac{280}{125}125280​

Correct answer: (C)

Step-by-step solution →
Q168·MathematicsNumericalJEE Main 2022
Let 3, 6, 9, 12,... upto 78 terms and 5, 9, 13, 17,... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to____.

Correct answer: 2223

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2022
Let {an}n=0∞\{a_{n}\}_{n=0}^{\infty}{an​}n=0∞​ be a sequence such that a0=a1=0a_{0} = a_{1} = 0a0​=a1​=0 and an+2=2an+1−an+1a_{n+2} = 2a_{n+1} - a_{n} + 1an+2​=2an+1​−an​+1 for all n≥0n \ge 0n≥0. Then, ∑n=2∞an7n\sum_{n=2}^{\infty}\frac{a_{n}}{7^{n}}∑n=2∞​7nan​​ is equal to
  1. (A)6343\frac{6}{343}3436​
  2. (B)7216\frac{7}{216}2167​
  3. (C)8343\frac{8}{343}3438​
  4. (D)49216\frac{49}{216}21649​

Correct answer: (B)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2022
Let A1,\mathrm{A}_{1},A1​, A2,\mathrm{A}_{2},A2​, A3,\mathrm{A}_{3},A3​, …… be an increasing geometric progression of positive real numbers. If A1A3A5A7=11296\mathrm{A}_{1}\mathrm{A}_{3}\mathrm{A}_{5}\mathrm{A}_{7}=\frac{1}{1296}A1​A3​A5​A7​=12961​ and A2+A4=736,\mathrm{A}_{2}+\mathrm{A}_{4}=\frac{7}{36},A2​+A4​=367​, then, the value of A6+A8+A10\mathrm{A}_{6}+\mathrm{A}_{8}+\mathrm{A}_{10}A6​+A8​+A10​ is equal to
  1. (A)33
  2. (B)37
  3. (C)43
  4. (D)47

Correct answer: (C)

Step-by-step solution →
Q171·MathematicsNumericalJEE Main 2022
Let for n=1,2,…,50n = 1, 2, \ldots, 50n=1,2,…,50, SnS_nSn​ be the sum of the infinite geometric progression whose first term is n2n^2n2 and whose common ratio is 1(n+1)2\frac{1}{(n+1)^2}(n+1)21​. Then the value of 126+∑n=150(Sn+2n+1−n−1)\frac{1}{26} + \sum\limits_{n=1}^{50}\left(S_n + \frac{2}{n+1} - n - 1\right)261​+n=1∑50​(Sn​+n+12​−n−1) is equal to

Correct answer: 41651

Step-by-step solution →
Q172·MathematicsSingle correctJEE Main 2022
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:71 : 71:7 and a+n=33a + n = 33a+n=33, then the value of n is
  1. (A)21
  2. (B)22
  3. (C)23
  4. (D)24

Correct answer: (C)

Step-by-step solution →
Q173·MathematicsNumericalJEE Main 2022
Let A={1, a1, a2……a18, 77}A = \{1,\, a_1,\, a_2 \ldots\ldots a_{18},\, 77\}A={1,a1​,a2​……a18​,77} be a set of integers with 1<a1<a2<…..<a18<771 < a_1 < a_2 < \ldots.. < a_{18} < 771<a1​<a2​<…..<a18​<77. Let the set A+A={x+y:x,y∈A}A + A = \{x + y : x, y \in A\}A+A={x+y:x,y∈A} contain exactly 39 elements. Then, the value of a1+a2+…..+a18a_1 + a_2 + \ldots.. + a_{18}a1​+a2​+…..+a18​ is equal to ______.

Correct answer: 702

Step-by-step solution →
Q174·MathematicsSingle correctJEE Main 2022
Let S=2+67+1272+2073+3074+.....S=2+\dfrac{6}{7}+\dfrac{12}{7^{2}}+\dfrac{20}{7^{3}}+\dfrac{30}{7^{4}}+.....S=2+76​+7212​+7320​+7430​+..... then 4S4S4S is equal to
  1. (A)(73)2\left(\dfrac{7}{3}\right)^{2}(37​)2
  2. (B)7332\dfrac{7^{3}}{3^{2}}3273​
  3. (C)(73)3\left(\dfrac{7}{3}\right)^{3}(37​)3
  4. (D)7233\dfrac{7^{2}}{3^{3}}3372​

Correct answer: (C)

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2022
If x=∑n=0∞anx = \sum_{n=0}^{\infty} a^{n}x=∑n=0∞​an, y=∑n=0∞bny = \sum_{n=0}^{\infty} b^{n}y=∑n=0∞​bn, z=∑n=0∞cnz = \sum_{n=0}^{\infty} c^{n}z=∑n=0∞​cn, where a, b, c are in A.P. and ∣a∣<1|a| < 1∣a∣<1, ∣b∣<1|b| < 1∣b∣<1, ∣c∣<1|c| < 1∣c∣<1, abc≠0abc \neq 0abc=0, then
  1. (A)x, y, z are in A.P.
  2. (B)x, y, z are in G.P.
  3. (C)1x,1y,1z\frac{1}{x}, \frac{1}{y}, \frac{1}{z}x1​,y1​,z1​ are in A.P.
  4. (D)1x+1y+1z=1−(a+b+c)\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - \left(a + b + c\right)x1​+y1​+z1​=1−(a+b+c)

Correct answer: (C)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2022
If a1,a2,a3....a_{1},a_{2},a_{3}....a1​,a2​,a3​.... and b1,b2,b3....b_{1},b_{2},b_{3}....b1​,b2​,b3​.... are A.P. and a1=2a_{1}=2a1​=2, a10=3a_{10}=3a10​=3, a1b1=1=a10b10a_{1}b_{1}=1=a_{10}b_{10}a1​b1​=1=a10​b10​ then a4b4a_{4}b_{4}a4​b4​ is equal to
  1. (A)3527\dfrac{35}{27}2735​
  2. (B)111
  3. (C)2728\dfrac{27}{28}2827​
  4. (D)2827\dfrac{28}{27}2728​

Correct answer: (D)

Step-by-step solution →
Q177·MathematicsNumericalJEE Main 2022
If the sum of the first ten terms of the series 15+265+3325+41025+52501+....\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+....51​+652​+3253​+10254​+25015​+.... is mn\frac{m}{n}nm​, where m and n are co-prime numbers, then m + n is equal to __________.

Correct answer: 276

Step-by-step solution →
Q178·MathematicsSingle correctJEE Main 2022
If A=∑n=1∞1(3+(−1)n)nA=\sum_{n=1}^{\infty}\frac{1}{\left(3+(-1)^n\right)^n}A=∑n=1∞​(3+(−1)n)n1​ and B=∑n=1∞(−1)n(3+(−1)n)nB=\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(3+(-1)^n\right)^n}B=∑n=1∞​(3+(−1)n)n(−1)n​, then AB\frac{A}{B}BA​ is equal to :
  1. (A)119\frac{11}{9}911​
  2. (B)1
  3. (C)−119-\frac{11}{9}−911​
  4. (D)−113-\frac{11}{3}−311​

Correct answer: (C)

Step-by-step solution →
Q179·MathematicsNumericalJEE Main 2022
If a1(>0)a_{1}(>0)a1​(>0), a2a_{2}a2​, a3a_{3}a3​, a4a_{4}a4​, a5a_{5}a5​ are in a G.P., a2+a4=2a3+1a_{2}+a_{4}=2a_{3}+1a2​+a4​=2a3​+1 and 3a2+a3=2a43a_{2}+a_{3}=2a_{4}3a2​+a3​=2a4​, then a2+a4+2a5a_{2}+a_{4}+2a_{5}a2​+a4​+2a5​ is equal to ____.

Correct answer: 40

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2022
Let A={n∈N:H.C.F. (n,45)=1}A = \{n \in N : \text{H.C.F. }(n, 45) = 1\}A={n∈N:H.C.F. (n,45)=1} and Let B={2k:k∈{1,2,...,100}}B = \{2k : k \in \{1, 2, ..., 100\}\}B={2k:k∈{1,2,...,100}}. Then the sum of all the elements of A∩BA \cap BA∩B is ___________.

Correct answer: 5264

Step-by-step solution →
Q181·MathematicsNumericalJEE Main 2022
Let A=∑i=110∑j=110min⁡{i,j}A = \sum_{i=1}^{10}\sum_{j=1}^{10}\min\{i, j\}A=∑i=110​∑j=110​min{i,j} and B=∑i=110∑j=110max⁡{i,j}B = \sum_{i=1}^{10}\sum_{j=1}^{10}\max\{i, j\}B=∑i=110​∑j=110​max{i,j}. Then A+BA + BA+B is equal to ________.

Correct answer: 1100

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2022
The sum 1+2⋅3+3⋅32+.....+10⋅391 + 2 \cdot 3 + 3 \cdot 3^2 + ..... + 10 \cdot 3^91+2⋅3+3⋅32+.....+10⋅39 is equal to
  1. (A)2⋅312+104\frac{2 \cdot 3^{12} + 10}{4}42⋅312+10​
  2. (B)19⋅310+14\frac{19 \cdot 3^{10} + 1}{4}419⋅310+1​
  3. (C)5⋅310−25 \cdot 3^{10} - 25⋅310−2
  4. (D)9⋅310+12\frac{9 \cdot 3^{10} + 1}{2}29⋅310+1​

Correct answer: (B)

Step-by-step solution →
Q183·MathematicsNumericalJEE Main 2022
The greatest integer less than or equal to the sum of first 100 terms of the sequence 13,59,1927,6581,……\frac{1}{3}, \frac{5}{9}, \frac{19}{27}, \frac{65}{81}, \ldots\ldots31​,95​,2719​,8165​,…… is equal to

Correct answer: 98

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2022
If 12⋅310+122⋅39+⋯1210⋅3=K210⋅310\frac{1}{2 \cdot 3^{10}} + \frac{1}{2^2 \cdot 3^9} + \cdots \frac{1}{2^{10} \cdot 3} = \frac{K}{2^{10} \cdot 3^{10}}2⋅3101​+22⋅391​+⋯210⋅31​=210⋅310K​, then the remainder when K is divided by 6 is
  1. (A)111
  2. (B)222
  3. (C)333
  4. (D)555

Correct answer: (D)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2022
Let f:N→Rf : N \to Rf:N→R be a function such that f(x+y)=2f(x)f(y)f(x+y)=2f(x)f(y)f(x+y)=2f(x)f(y) for natural numbers x and y. If f(1)=2f(1) = 2f(1)=2, then the value of α for which ∑k=110f(α+k)=5123(220−1)\sum_{k=1}^{10} f(\alpha + k) = \frac{512}{3}(2^{20} - 1)∑k=110​f(α+k)=3512​(220−1) holds, is
  1. (A)222
  2. (B)333
  3. (C)444
  4. (D)666

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsNumericalJEE Main 2022
For a natural number n, let an=19n−12na_n = 19^n - 12^nan​=19n−12n. Then, the value of 31α9−α1057α8\frac{31\alpha_9 - \alpha_{10}}{57\alpha_8}57α8​31α9​−α10​​ is

Correct answer: 4

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2022
Let x,y>0x, y > 0x,y>0. If x3y2=215x^{3}y^{2} = 2^{15}x3y2=215, then the least value of 3x+2y3x + 2y3x+2y is
  1. (A)30
  2. (B)32
  3. (C)36
  4. (D)40

Correct answer: (D)

Step-by-step solution →
Q188·MathematicsNumericalJEE Main 2022
The sum of all the elements of the set {α∈{1,2,…,100}:HCF (α,24)=1}\{\alpha \in \{1, 2, \ldots, 100\} : \mathrm{HCF}\,(\alpha, 24) = 1\}{α∈{1,2,…,100}:HCF(α,24)=1} is ______.

Correct answer: 1633

Step-by-step solution →
Q189·MathematicsSingle correctJEE Main 2022
If {ai}i=1n\{a_i\}_{i=1}^{n}{ai​}i=1n​ where n is an even integer , is an arithmetic progression with common difference 1, and ∑i=1nai=192\sum_{i=1}^{n} a_i = 192∑i=1n​ai​=192, ∑i=1n/2a2i=120\sum_{i=1}^{n/2} a_{2i} = 120∑i=1n/2​a2i​=120, then n is equal to:
  1. (A)48
  2. (B)96
  3. (C)92
  4. (D)104

Correct answer: (B)

Step-by-step solution →
Q190·MathematicsSingle correctJEE Advanced 2021
Let M={(x,y)∈R×R:x2+y2≤r2},M = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \le r^2 \right\},M={(x,y)∈R×R:x2+y2≤r2}, where r>0r > 0r>0. Consider the geometric progression an=12n−1a_n = \frac{1}{2^{n-1}}an​=2n−11​, n=1,2,3,…n = 1, 2, 3, \ldotsn=1,2,3,… . Let S0=0S_0 = 0S0​=0 and, for n≥1n \ge 1n≥1, let SnS_nSn​ denote the sum of the first n terms of this progression. For n≥1n \ge 1n≥1, let CnC_nCn​ denote the circle with center (Sn−1,0)(S_{n-1}, 0)(Sn−1​,0) and radius ana_nan​, and DnD_nDn​ denote the circle with center (Sn−1,Sn−1)(S_{n-1}, S_{n-1})(Sn−1​,Sn−1​) and radius ana_nan​. Consider M with r=(2199−1)22198r = \frac{(2^{199} - 1)\sqrt{2}}{2^{198}}r=2198(2199−1)2​​. The number of all those circles DnD_nDn​ that are inside M is
  1. (A)198
  2. (B)199
  3. (C)200
  4. (D)201

Correct answer: (B)

Step-by-step solution →
Q191·MathematicsSingle correctJEE Advanced 2021
Let M={(x,y)∈R×R:x2+y2≤r2},M = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \le r^2 \right\},M={(x,y)∈R×R:x2+y2≤r2}, where r>0r > 0r>0. Consider the geometric progression an=12n−1a_n = \frac{1}{2^{n-1}}an​=2n−11​, n=1,2,3,…n = 1, 2, 3, \ldotsn=1,2,3,… . Let S0=0S_0 = 0S0​=0 and, for n≥1n \ge 1n≥1, let SnS_nSn​ denote the sum of the first n terms of this progression. For n≥1n \ge 1n≥1, let CnC_nCn​ denote the circle with center (Sn−1,0)(S_{n-1}, 0)(Sn−1​,0) and radius ana_nan​, and DnD_nDn​ denote the circle with center (Sn−1,Sn−1)(S_{n-1}, S_{n-1})(Sn−1​,Sn−1​) and radius ana_nan​. Consider M with r=1025513r = \frac{1025}{513}r=5131025​. Let k be the number of all those circles CnC_nCn​ that are inside M. Let lll be the maximum possible number of circles among these k circles such that no two circles intersect. Then
  1. (A)k+2l=22k + 2l = 22k+2l=22
  2. (B)2k+l=262k + l = 262k+l=26
  3. (C)2k+3l=342k + 3l = 342k+3l=34
  4. (D)3k+2l=403k + 2l = 403k+2l=40

Correct answer: (D)

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2021
Let a1,a2,.......,a21a_{1}, a_{2}, ......., a_{21}a1​,a2​,.......,a21​ be an AP such that ∑n=1201anan+1=49\sum_{n=1}^{20}\frac{1}{a_{n}a_{n+1}} = \frac{4}{9}∑n=120​an​an+1​1​=94​. If the sum of this AP is 189, then a6a16a_{6}a_{16}a6​a16​ is equal to :
  1. (A)57
  2. (B)72
  3. (C)48
  4. (D)36

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2021
Let Sn=1⋅(n−1)+2⋅(n−2)+3⋅(n−3)+⋯+(n−1)⋅1S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1Sn​=1⋅(n−1)+2⋅(n−2)+3⋅(n−3)+⋯+(n−1)⋅1 , n ≥ 4. The sum ∑n=4∞(2Snn!−1(n−2)!)\sum_{n=4}^{\infty}\left(\frac{2S_{n}}{n!} - \frac{1}{(n-2)!}\right)∑n=4∞​(n!2Sn​​−(n−2)!1​) is equal to :
  1. (A)e−13\frac{e-1}{3}3e−1​
  2. (B)e−26\frac{e-2}{6}6e−2​
  3. (C)e3\frac{e}{3}3e​
  4. (D)e6\frac{e}{6}6e​

Correct answer: (A)

Step-by-step solution →
Q194·MathematicsNumericalJEE Main 2021
If S=75+952+1353+1954+....S = \frac{7}{5} + \frac{9}{5^{2}} + \frac{13}{5^{3}} + \frac{19}{5^{4}} + ....S=57​+529​+5313​+5419​+...., then 160 S is equal to ______.

Correct answer: 305

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2021
Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3r23r^{2}3r2, then r2−dr^{2} - dr2−d is equal to :
  1. (A)7−737 - 7\sqrt{3}7−73​
  2. (B)7+37 + \sqrt{3}7+3​
  3. (C)7−37 - \sqrt{3}7−3​
  4. (D)7+337 + 3\sqrt{3}7+33​

Correct answer: (B)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2021
The sum of 10 terms of the series 312×22+522×32+732×42+....\frac{3}{1^{2} \times 2^{2}} + \frac{5}{2^{2} \times 3^{2}} + \frac{7}{3^{2} \times 4^{2}} + ....12×223​+22×325​+32×427​+.... is :
  1. (A)1
  2. (B)120121\frac{120}{121}121120​
  3. (C)99100\frac{99}{100}10099​
  4. (D)143144\frac{143}{144}144143​

Correct answer: (B)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2021
Let a1,a2,a3,…a_1, a_2, a_3, \ldotsa1​,a2​,a3​,… be an A.P. If a1+a2+…+a10a1+a2+…+ap=100p2\frac{a_1 + a_2 + \ldots + a_{10}}{a_1 + a_2 + \ldots + a_p} = \frac{100}{p^2}a1​+a2​+…+ap​a1​+a2​+…+a10​​=p2100​, p≠10p \neq 10p=10, then a11a10\frac{a_{11}}{a_{10}}a10​a11​​ is equal to :
  1. (A)1921\frac{19}{21}2119​
  2. (B)100121\frac{100}{121}121100​
  3. (C)2119\frac{21}{19}1921​
  4. (D)121100\frac{121}{100}100121​

Correct answer: (C)

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2021
If 0 < x < 1 and y=12x2+23x3+34x4+…y = \frac{1}{2}x^{2} + \frac{2}{3}x^{3} + \frac{3}{4}x^{4} + \ldotsy=21​x2+32​x3+43​x4+…, then the value of e1+ye^{1+y}e1+y at x=12x = \frac{1}{2}x=21​ is:
  1. (A)12e2\frac{1}{2}e^{2}21​e2
  2. (B)2e
  3. (C)12e\frac{1}{2}\sqrt{e}21​e​
  4. (D)2e22e^{2}2e2

Correct answer: (A)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2021
If 0<x<10 < x < 10<x<1, then 32x2+53x3+74x4+.....\frac{3}{2}x^2 + \frac{5}{3}x^3 + \frac{7}{4}x^4 + .....23​x2+35​x3+47​x4+..... , is equal to :
  1. (A)x(1+x1−x)+log⁡e(1−x)x\left(\frac{1+x}{1-x}\right) + \log_e(1-x)x(1−x1+x​)+loge​(1−x)
  2. (B)x(1−x1+x)+log⁡e(1−x)x\left(\frac{1-x}{1+x}\right) + \log_e(1-x)x(1+x1−x​)+loge​(1−x)
  3. (C)1−x1+x+log⁡e(1−x)\frac{1-x}{1+x} + \log_e(1-x)1+x1−x​+loge​(1−x)
  4. (D)1+x1−x+log⁡e(1−x)\frac{1+x}{1-x} + \log_e(1-x)1−x1+x​+loge​(1−x)

Correct answer: (A)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2021
If for x,y∈Rx, y \in \mathbf{R}x,y∈R, x>0x > 0x>0, y=log⁡10x+log⁡10x1/3+log⁡10x1/9+.....y = \log_{10}x + \log_{10}x^{1/3} + \log_{10}x^{1/9} + .....y=log10​x+log10​x1/3+log10​x1/9+..... upto ∞\infty∞ terms and 2+4+6+....+2y3+6+9+....+3y=4log⁡10x\frac{2+4+6+....+2y}{3+6+9+....+3y} = \frac{4}{\log_{10} x}3+6+9+....+3y2+4+6+....+2y​=log10​x4​ , then the ordered pair (x,y)(x, y)(x,y) is equal to :
  1. (A)(106,6)(10^6, 6)(106,6)
  2. (B)(104,6)(10^4, 6)(104,6)
  3. (C)(102,3)(10^2, 3)(102,3)
  4. (D)(106,9)(10^6, 9)(106,9)

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2021
If the sum of an infinite GP a, ar, ar2ar^2ar2, ar3ar^3ar3,... is 15 and the sum of the squares of its each term is 150, then the sum of ar2ar^2ar2, ar4ar^4ar4, ar6ar^6ar6, ... is :
  1. (A)52\frac{5}{2}25​
  2. (B)12\frac{1}{2}21​
  3. (C)252\frac{25}{2}225​
  4. (D)92\frac{9}{2}29​

Correct answer: (B)

Step-by-step solution →
Q202·MathematicsNumericalJEE Main 2021
Let a1,a2,.....,a10a_{1}, a_{2}, ....., a_{10}a1​,a2​,.....,a10​ be an AP with common difference −3-3−3 and b1,b2,.....,b10b_{1}, b_{2}, ....., b_{10}b1​,b2​,.....,b10​ be a GP with common ratio 2. Let ck=ak+bkc_{k} = a_{k} + b_{k}ck​=ak​+bk​, k=1,2,...,10k = 1, 2, ..., 10k=1,2,...,10. If c2=12c_{2} = 12c2​=12 and c3=13c_{3} = 13c3​=13, then ∑k=110ck\sum_{k=1}^{10} c_{k}∑k=110​ck​ is equal to ________.

Correct answer: 2021

Step-by-step solution →
Q203·MathematicsNumericalJEE Main 2021
If log⁡32,log⁡3(2x−5),log⁡3(2x−72)\log_3 2, \log_3\left(2^x-5\right), \log_3\left(2^x-\frac{7}{2}\right)log3​2,log3​(2x−5),log3​(2x−27​) are in an arithmetic progression, then the value of x is equal to ____.

Correct answer: 3

Step-by-step solution →
Q204·MathematicsSingle correctJEE Main 2021
Let SnS_{n}Sn​ be the sum of the first n terms of an arithmetic progression. If S3n=3S2nS_{3n}=3S_{2n}S3n​=3S2n​, then the value of S4nS2n\frac{S_{4n}}{S_{2n}}S2n​S4n​​ is ;
  1. (A)2
  2. (B)6
  3. (C)8
  4. (D)4

Correct answer: (B)

Step-by-step solution →
Q205·MathematicsNumericalJEE Main 2021
If the value of (1+23+632+1033+....upto ∞)log⁡0.25(13+132+....upto ∞)\left( 1 + \dfrac{2}{3} + \dfrac{6}{3^{2}} + \dfrac{10}{3^{3}} + .... \text{upto } \infty \right)^{\log_{0.25}\left( \dfrac{1}{3} + \dfrac{1}{3^{2}} + .... \text{upto } \infty \right)}(1+32​+326​+3310​+....upto ∞)log0.25​(31​+321​+....upto ∞) is l, then l2l^{2}l2 is equal to ........

Correct answer: 3

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2021
Let SnS_nSn​ denote the sum of the first n-terms of an arithmetic progression. If S10=530S_{10} = 530S10​=530, S5=140S_5 = 140S5​=140, then S20−S6S_{20} - S_6S20​−S6​ is equal to :
  1. (A)185218521852
  2. (B)184218421842
  3. (C)187218721872
  4. (D)186218621862

Correct answer: (D)

Step-by-step solution →
Q207·MathematicsNumericalJEE Main 2021
Let {an}n=1∞\{a_{n}\}_{n=1}^{\infty}{an​}n=1∞​ be a sequence such that a1=1,a2=1a_{1}=1, a_{2}=1a1​=1,a2​=1 and an+2=2an+1+ana_{n+2}=2a_{n+1}+a_{n}an+2​=2an+1​+an​ for all n≥1n\ge 1n≥1. Then the value of 47∑n=1∞an23n47\sum_{n=1}^{\infty}\frac{a_{n}}{2^{3n}}47∑n=1∞​23nan​​ is equal to……..

Correct answer: 7

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2021
If sum of the first 21 terms of the series log⁡91/2x+log⁡91/3x+log⁡91/4x+....,\log_{9^{1/2}}x+\log_{9^{1/3}}x+\log_{9^{1/4}}x+....,log91/2​x+log91/3​x+log91/4​x+...., where x>0x>0x>0 is 504, then x is equal to :
  1. (A)777
  2. (B)999
  3. (C)243243243
  4. (D)818181

Correct answer: (D)

Step-by-step solution →
Q209·MathematicsNumericalJEE Main 2021
If ∑r=110r!(r3+6r2+2r+5)=α(11!)\sum_{r=1}^{10} r!\left(r^{3} + 6r^{2} + 2r + 5\right) = \alpha(11!)∑r=110​r!(r3+6r2+2r+5)=α(11!), then the value of α is equal to ________ .

Correct answer: 160

Step-by-step solution →
Q210·MathematicsSingle correctJEE Main 2021
Let S1S_{1}S1​ be the sum of first 2n terms of an arithmetic progression. Let S2S_{2}S2​ be the sum of first 4n terms of the same arithmetic progression. If (S2−S1)(S_{2} - S_{1})(S2​−S1​) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:
  1. (A)1000
  2. (B)7000
  3. (C)5000
  4. (D)3000

Correct answer: (D)

Step-by-step solution →
Q211·MathematicsSingle correctJEE Main 2021
132−1+152−1+172−1+...+1(201)2−1\dfrac{1}{3^{2}-1} + \dfrac{1}{5^{2}-1} + \dfrac{1}{7^{2}-1} + ... + \dfrac{1}{\left(201\right)^{2}-1}32−11​+52−11​+72−11​+...+(201)2−11​ is equal to
  1. (A)101404\dfrac{101}{404}404101​
  2. (B)25101\dfrac{25}{101}10125​
  3. (C)101408\dfrac{101}{408}408101​
  4. (D)99400\dfrac{99}{400}40099​

Correct answer: (B)

Step-by-step solution →
Q212·MathematicsNumericalJEE Main 2021
The missing value in the following figure is

Correct answer: 4

Step-by-step solution →
Q213·MathematicsSingle correctJEE Main 2021
If α,β\alpha, \betaα,β are natural numbers such that 100α−199β=(100)(100)+(99)(101)+(98)(102)+.....+(1)(199)100^{\alpha} - 199\beta = (100)(100) + (99)(101) + (98)(102) + ..... + (1)(199)100α−199β=(100)(100)+(99)(101)+(98)(102)+.....+(1)(199), then the slope of the line passing through (α,β)(\alpha, \beta)(α,β) and origin is :
  1. (A)540540540
  2. (B)550550550
  3. (C)530530530
  4. (D)510510510

Correct answer: (B)

Step-by-step solution →
Q214·MathematicsSingle correctJEE Main 2021
The value of 4+15+14+15+14+.......∞4 + \cfrac{1}{5 + \cfrac{1}{4 + \cfrac{1}{5 + \cfrac{1}{4 + .......\infty}}}}4+5+4+5+4+.......∞1​1​1​1​ is :
  1. (A)2+25302 + \frac{2}{5}\sqrt{30}2+52​30​
  2. (B)2+45302 + \frac{4}{\sqrt{5}}\sqrt{30}2+5​4​30​
  3. (C)4+45304 + \frac{4}{\sqrt{5}}\sqrt{30}4+5​4​30​
  4. (D)5+25305 + \frac{2}{5}\sqrt{30}5+52​30​

Correct answer: (A)

Step-by-step solution →
Q215·MathematicsNumericalJEE Main 2021
Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ______ .

Correct answer: 3

Step-by-step solution →
Q216·MathematicsNumericalJEE Main 2021
Let Sn(x)=log⁡a1/2x+log⁡a1/3x+log⁡a1/6xS_{n}(x)=\log_{a^{1/2}}x+\log_{a^{1/3}}x+\log_{a^{1/6}}xSn​(x)=loga1/2​x+loga1/3​x+loga1/6​x +log⁡a1/11x+log⁡a1/18x+log⁡a1/27x+.....+\log_{a^{1/11}}x+\log_{a^{1/18}}x+\log_{a^{1/27}}x+.....+loga1/11​x+loga1/18​x+loga1/27​x+..... up to n-terms, where a > 1. If S24(x)=1093S_{24}(x) = 1093S24​(x)=1093 and S12(2x)=265S_{12}(2x) = 265S12​(2x)=265, then value of a is equal to _______.

Correct answer: 16

Step-by-step solution →
Q217·MathematicsNumericalJEE Main 2021
Let 116\frac{1}{16}161​, a and b be in G.P. and 1a,1b,6\frac{1}{a},\frac{1}{b},6a1​,b1​,6 be in A.P., where a, b > 0. Then 72(a+b)72(a + b)72(a+b) is equal to _________.

Correct answer: 14

Step-by-step solution →
Q218·MathematicsSingle correctJEE Main 2021
The sum of the infinite series 1+23+732+1233+1734+2235+…1 + \frac{2}{3} + \frac{7}{3^{2}} + \frac{12}{3^{3}} + \frac{17}{3^{4}} + \frac{22}{3^{5}} + \ldots1+32​+327​+3312​+3417​+3522​+… is equal to
  1. (A)94\frac{9}{4}49​
  2. (B)154\frac{15}{4}415​
  3. (C)134\frac{13}{4}413​
  4. (D)114\frac{11}{4}411​

Correct answer: (C)

Step-by-step solution →
Q219·MathematicsSingle correctJEE Main 2021
In an increasing, geometric series, the sum of the second and the sixth term is 252\frac{25}{2}225​ and the product of the third and fifth term is 25. Then, the sum of 4th4^{th}4th, 6th6^{th}6th and 8th8^{th}8th terms is equal to :
  1. (A)35
  2. (B)30
  3. (C)26
  4. (D)32

Correct answer: (A)

Step-by-step solution →
Q220·MathematicsSingle correctJEE Main 2021
The sum of the series ∑n=1∞n2+6n+10(2n+1)!\sum_{n=1}^{\infty} \frac{n^{2} + 6n + 10}{(2n+1)!}∑n=1∞​(2n+1)!n2+6n+10​ is equal to :
  1. (A)418e+198e−1−10\frac{41}{8} e + \frac{19}{8} e^{-1} - 10841​e+819​e−1−10
  2. (B)−418e+198e−1−10-\frac{41}{8} e + \frac{19}{8} e^{-1} - 10−841​e+819​e−1−10
  3. (C)418e−198e−1−10\frac{41}{8} e - \frac{19}{8} e^{-1} - 10841​e−819​e−1−10
  4. (D)418e+198e−1+10\frac{41}{8} e + \frac{19}{8} e^{-1} + 10841​e+819​e−1+10

Correct answer: (C)

Step-by-step solution →
Q221·MathematicsNumericalJEE Main 2021
If the arithmetic mean and geometric mean of the pthp^{th}pth and qthq^{th}qth terms of the sequence −16-16−16, 8, −4-4−4, 2, ………… satisfy the equation 4x2−9x+5=04x^{2} - 9x + 5 = 04x2−9x+5=0, then p+q is equal to ________________.

Correct answer: 10

Step-by-step solution →
Q222·MathematicsNumericalJEE Main 2021
Let A1,A2,A3,…A_1, A_2, A_3, \dotsA1​,A2​,A3​,… be squares such that for each n≥1n \ge 1n≥1, the length of the side of AnA_nAn​ equals the length of diagonal of An+1A_{n+1}An+1​. If the length of A1A_1A1​ is 12 cm, then the smallest value of nnn for which area of AnA_nAn​ is less than one, is __________.

Correct answer: 9

Step-by-step solution →
Q223·MathematicsSingle correctJEE Main 2021
If 0<θ,ϕ<π20 < \theta, \phi < \frac{\pi}{2}0<θ,ϕ<2π​, x=∑n=0∞cos⁡2nθx = \sum_{n=0}^{\infty} \cos^{2n}\thetax=∑n=0∞​cos2nθ, y=∑n=0∞sin⁡2nϕy = \sum_{n=0}^{\infty} \sin^{2n}\phiy=∑n=0∞​sin2nϕ and z=∑n=0∞cos⁡2nθ⋅sin⁡2nϕz = \sum_{n=0}^{\infty} \cos^{2n}\theta \cdot \sin^{2n}\phiz=∑n=0∞​cos2nθ⋅sin2nϕ then :
  1. (A)xyz=4xyz = 4xyz=4
  2. (B)xy−z=(x+y)zxy - z = (x + y)zxy−z=(x+y)z
  3. (C)xy+yz+zx=zxy + yz + zx = zxy+yz+zx=z
  4. (D)xy+z=(x+y)zxy + z = (x + y)zxy+z=(x+y)z

Correct answer: (D)

Step-by-step solution →
Q224·MathematicsNumericalJEE Main 2021
The sum of first four terms of a geometric progression (G.P.) is 6512\frac{65}{12}1265​ and the sum of their respective reciprocals is 6518\frac{65}{18}1865​. If the product of first three terms of the G.P. is 1, and the third term is α\alphaα, then 2α2\alpha2α is__________.

Correct answer: 3

Step-by-step solution →
Q225·MathematicsNumericalJEE Main 2021
Let A={n∈N:n is a 3-digit number}A = \{n \in N : n \text{ is a 3-digit number}\}A={n∈N:n is a 3-digit number} B={9k+2:k∈N}B = \{9k + 2 : k \in N\}B={9k+2:k∈N} and C:{9k+ℓ:k∈N}C : \{9k + \ell : k \in N\}C:{9k+ℓ:k∈N} for some ℓ\ellℓ (0<ℓ<9)(0 < \ell < 9)(0<ℓ<9) If the sum of all the elements of the set A∩(B∪C)A \cap (B \cup C)A∩(B∪C) is 274×400274 \times 400274×400, then ℓ\ellℓ is equal to __

Correct answer: 5

Step-by-step solution →
Q226·MathematicsNumericalJEE Advanced 2020
Let a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​, ..... be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1b_{1}b1​, b2b_{2}b2​, b3b_{3}b3​, ..... be a sequence of positive integers in geometric progression with common ratio 2. If a1=b1=ca_{1} = b_{1} = ca1​=b1​=c, then the number of all possible values of c, for which the equality 2(a1+a2+…+an)=b1+b2+…+bn2\left(a_{1} + a_{2} + \ldots + a_{n}\right) = b_{1} + b_{2} + \ldots + b_{n}2(a1​+a2​+…+an​)=b1​+b2​+…+bn​ holds for some positive integer n, is ______

Correct answer: 1.00

Step-by-step solution →
Q227·MathematicsNumericalJEE Advanced 2020
Let m be the minimum possible value of log⁡3(3y1+3y2+3y3)\log_{3}\left(3^{y_{1}} + 3^{y_{2}} + 3^{y_{3}}\right)log3​(3y1​+3y2​+3y3​), where y1,y2,y3y_{1}, y_{2}, y_{3}y1​,y2​,y3​ are real numbers for which y1+y2+y3=9y_{1} + y_{2} + y_{3} = 9y1​+y2​+y3​=9. Let M be the maximum possible value of (log⁡3x1+log⁡3x2+log⁡3x3)\left(\log_{3} x_{1} + \log_{3} x_{2} + \log_{3} x_{3}\right)(log3​x1​+log3​x2​+log3​x3​), where x1,x2,x3x_{1}, x_{2}, x_{3}x1​,x2​,x3​ are positive real numbers for which x1+x2+x3=9x_{1} + x_{2} + x_{3} = 9x1​+x2​+x3​=9. Then the value of log⁡2(m3)+log⁡3(M2)\log_{2}\left(m^{3}\right) + \log_{3}\left(M^{2}\right)log2​(m3)+log3​(M2) is ______.

Correct answer: 8.00

Step-by-step solution →
Q228·MathematicsSingle correctJEE Main 2020
Let a, b, c, d and p be any non zero distinct real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0(a^{2}+b^{2}+c^{2})p^{2}-2(ab+bc+cd)p+(b^{2}+c^{2}+d^{2})=0(a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)=0. Then:
  1. (A)a, c, p are in A.P.
  2. (B)a, c, p are in G.P.
  3. (C)a, b, c, d are in G.P.
  4. (D)a, b, c, d are in A.P.

Correct answer: (C)

Step-by-step solution →
Q229·MathematicsSingle correctJEE Main 2020
If f(x+y)=f(x)f(y)f(x+y) = f(x)f(y)f(x+y)=f(x)f(y) and ∑x=1∞f(x)=2, x,y∈N\displaystyle\sum_{x=1}^{\infty} f(x) = 2,\ x, y \in Nx=1∑∞​f(x)=2, x,y∈N where N is the set of all natural numbers, then the value of f(4)f(2)\dfrac{f(4)}{f(2)}f(2)f(4)​ is:
  1. (A)23\dfrac{2}{3}32​
  2. (B)19\dfrac{1}{9}91​
  3. (C)13\dfrac{1}{3}31​
  4. (D)49\dfrac{4}{9}94​

Correct answer: (D)

Step-by-step solution →
Q230·MathematicsSingle correctJEE Main 2020
The common difference of the A.P. b1,b2,…,bmb_{1}, b_{2}, \ldots, b_{m}b1​,b2​,…,bm​ is 2 more than the common difference of A.P. a1,a2,…,ana_{1}, a_{2}, \ldots, a_{n}a1​,a2​,…,an​. If a40=−159a_{40} = -159a40​=−159, a100=−399a_{100} = -399a100​=−399 and b100=a70b_{100} = a_{70}b100​=a70​, then b1b_{1}b1​ is equal to:
  1. (A)81
  2. (B)−127
  3. (C)−81
  4. (D)127

Correct answer: (C)

Step-by-step solution →
Q231·MathematicsSingle correctJEE Main 2020
If 210+29.31+28.32+...+2.39+310=S−2112^{10} + 2^{9}.3^{1} + 2^{8}.3^{2} + ... + 2.3^{9} + 3^{10} = S - 2^{11}210+29.31+28.32+...+2.39+310=S−211, then S is equal to
  1. (A)3112+210\dfrac{3^{11}}{2} + 2^{10}2311​+210
  2. (B)311−2123^{11} - 2^{12}311−212
  3. (C)2.3112.3^{11}2.311
  4. (D)3113^{11}311

Correct answer: (D)

Step-by-step solution →
Q232·MathematicsSingle correctJEE Main 2020
If the sum of the first 20 terms of the series log⁡(71/2)x+log⁡(71/3)x+log⁡(71/4)x+...\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + ...log(71/2)​x+log(71/3)​x+log(71/4)​x+... is 460, then x is equal to:
  1. (A)727^272
  2. (B)746/217^{46/21}746/21
  3. (C)71/27^{1/2}71/2
  4. (D)e2e^2e2

Correct answer: (A)

Step-by-step solution →
Q233·MathematicsSingle correctJEE Main 2020
If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eights terms is 243, then the sum of the first 50 terms of this G.P. is:
  1. (A)126(350−1)\dfrac{1}{26}\left(3^{50}-1\right)261​(350−1)
  2. (B)113(350−1)\dfrac{1}{13}\left(3^{50}-1\right)131​(350−1)
  3. (C)213(350−1)\dfrac{2}{13}\left(3^{50}-1\right)132​(350−1)
  4. (D)126(349−1)\dfrac{1}{26}\left(3^{49}-1\right)261​(349−1)

Correct answer: (A)

Step-by-step solution →
Q234·MathematicsSingle correctJEE Main 2020
If 32sin⁡2α−13^{2\sin 2\alpha - 1}32sin2α−1, 14 and 34−2sin⁡2α3^{4-2\sin 2\alpha}34−2sin2α are the first three terms of an AP for some α\alphaα, then the sixth term of this AP is:
  1. (A)65
  2. (B)78
  3. (C)66
  4. (D)81

Correct answer: (C)

Step-by-step solution →
Q235·MathematicsSingle correctJEE Main 2020
If 1+(1−22⋅1)+(1−42⋅3)+(1−62⋅5)+……+(1−202⋅19)=α−220β1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + \ldots\ldots + (1 - 20^{2} \cdot 19) = \alpha - 220\beta1+(1−22⋅1)+(1−42⋅3)+(1−62⋅5)+……+(1−202⋅19)=α−220β, then an ordered pair (α,β)(\alpha, \beta)(α,β) is equal to:
  1. (A)(11,97)(11, 97)(11,97)
  2. (B)(10,97)(10, 97)(10,97)
  3. (C)(10,103)(10, 103)(10,103)
  4. (D)(11,103)(11, 103)(11,103)

Correct answer: (D)

Step-by-step solution →
Q236·MathematicsSingle correctJEE Main 2020
If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is:
  1. (A)14\frac{1}{4}41​
  2. (B)17\frac{1}{7}71​
  3. (C)16\frac{1}{6}61​
  4. (D)15\frac{1}{5}51​

Correct answer: (C)

Step-by-step solution →
Q237·MathematicsNumericalJEE Main 2020
If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th4^{th}4th A.M. is equal to 2nd2^{nd}2nd G.M., then m is equal to ________________.

Correct answer: 39.00

Step-by-step solution →
Q238·MathematicsNumericalJEE Main 2020
The value of (0.16)log⁡2.5(13+132+133+… to ∞)(0.16)^{\log_{2.5}\left(\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \ldots \text{ to } \infty\right)}(0.16)log2.5​(31​+321​+331​+… to ∞) is equal to __________.

Correct answer: 4

Step-by-step solution →
Q239·MathematicsSingle correctJEE Main 2020
If the sum of the series 20+1935+1915+1845+....20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + ....20+1953​+1951​+1854​+.... upto nthn^{th}nth term is 488 and the nthn^{th}nth term is negative, then:
  1. (A)nthn^{th}nth term is −425-4\frac{2}{5}−452​
  2. (B)n=41n = 41n=41
  3. (C)nthn^{th}nth term is −4-4−4
  4. (D)n=60n = 60n=60

Correct answer: (C)

Step-by-step solution →
Q240·MathematicsNumericalJEE Main 2020
The number of terms common to the two A.P.'s 3, 7, 11,…………..407 and 2, 9, 16, …..709 is ______

Correct answer: 14

Step-by-step solution →
Q241·MathematicsSingle correctJEE Main 2020
The product 214.4116.8148.1611282^{\frac{1}{4}}.4^{\frac{1}{16}}.8^{\frac{1}{48}}.16^{\frac{1}{128}}241​.4161​.8481​.161281​.......... to ∞\infty∞ is equal to
  1. (A)2122^{\frac{1}{2}}221​
  2. (B)2
  3. (C)1
  4. (D)2142^{\frac{1}{4}}241​

Correct answer: (A)

Step-by-step solution →
Q242·MathematicsSingle correctJEE Main 2020
If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n=0}^{\infty} (-1)^{n} \tan^{2n}\thetax=n=0∑∞​(−1)ntan2nθ and y=∑n=0∞cos⁡2nθy = \sum\limits_{n=0}^{\infty} \cos^{2n}\thetay=n=0∑∞​cos2nθ, for 0<θ<π40 < \theta < \dfrac{\pi}{4}0<θ<4π​, then:
  1. (A)y(1−x)=1y(1 - x) = 1y(1−x)=1
  2. (B)y(1+x)=1y(1 + x) = 1y(1+x)=1
  3. (C)x(1+y)=1x(1 + y) = 1x(1+y)=1
  4. (D)x(1−y)=1x(1 - y) = 1x(1−y)=1

Correct answer: (A)

Step-by-step solution →
Q243·MathematicsSingle correctJEE Main 2020
Let ana_{n}an​ be the nthn^{th}nth term of a G.P. of positive terms. If ∑n=1100a2n+1=200\sum\limits_{n=1}^{100} a_{2n+1} = 200n=1∑100​a2n+1​=200 and ∑n=1100a2n=100\sum\limits_{n=1}^{100} a_{2n} = 100n=1∑100​a2n​=100, then ∑n=1200an\sum\limits_{n=1}^{200} a_{n}n=1∑200​an​ is equal to:
  1. (A)225
  2. (B)300
  3. (C)150
  4. (D)175

Correct answer: (C)

Step-by-step solution →
Q244·MathematicsSingle correctJEE Main 2020
Let f:R→Rf:R\to Rf:R→R be such that for all x∈Rx\in Rx∈R, (21+x+21−x),f(x)\left(2^{1+x}+2^{1-x}\right), f(x)(21+x+21−x),f(x) and (3x+3−x)\left(3^{x}+3^{-x}\right)(3x+3−x) are in A.P., then the minimum value of f(x)f(x)f(x) is:
  1. (A)333
  2. (B)444
  3. (C)222
  4. (D)000

Correct answer: (A)

Step-by-step solution →
Q245·MathematicsSingle correctJEE Main 2020
If the 10th10^{th}10th term of an A.P. is 120\dfrac{1}{20}201​ and its 20th20^{th}20th term is 110\dfrac{1}{10}101​, then the sum of its first 200 terms is:
  1. (A)100
  2. (B)501450\dfrac{1}{4}5041​
  3. (C)10012100\dfrac{1}{2}10021​
  4. (D)50

Correct answer: (C)

Step-by-step solution →
Q246·MathematicsNumericalJEE Main 2020
The sum ∑k=120(1+2+3+....+k)\displaystyle\sum_{k=1}^{20}(1+2+3+....+k)k=1∑20​(1+2+3+....+k) is

Correct answer: 1540

Step-by-step solution →
Q247·MathematicsNumericalJEE Main 2020
The sum ∑n=17n(n+1)(2n+1)4\sum_{n=1}^{7}\dfrac{n(n+1)(2n+1)}{4}∑n=17​4n(n+1)(2n+1)​ is equal to ___________.

Correct answer: 504

Step-by-step solution →
Q248·MathematicsSingle correctJEE Main 2020
If the sum of the first 40 terms of the series, 3+4+8+9+13+14+18+19+....3+4+8+9+13+14+18+19+....3+4+8+9+13+14+18+19+.... is (102)m, then m is equal to:
  1. (A)10
  2. (B)5
  3. (C)20
  4. (D)25

Correct answer: (C)

Step-by-step solution →
Q249·MathematicsSingle correctJEE Main 2020
Let a1,a2,a3,........a_1,a_2,a_3,........a1​,a2​,a3​,........ be a G.P. such that a1<0,a1+a2=4a_1<0, a_1+a_2=4a1​<0,a1​+a2​=4 and a3+a4=16a_3+a_4=16a3​+a4​=16. If ∑i=19ai=4λ\displaystyle\sum_{i=1}^{9}a_i=4\lambdai=1∑9​ai​=4λ, then λ\lambdaλ is equal to:
  1. (A)-171
  2. (B)-513
  3. (C)171
  4. (D)5113\dfrac{511}{3}3511​

Correct answer: (A)

Step-by-step solution →
Q250·MathematicsSingle correctJEE Main 2020
Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is −12-\frac{1}{2}−21​, then the greatest number amongst them is:
  1. (A)7
  2. (B)212\frac{21}{2}221​
  3. (C)16
  4. (D)27

Correct answer: (C)

Step-by-step solution →
Q251·MathematicsSingle correctJEE Main 2020
The greatest positive integer k, for which 49k+149^{k}+149k+1 is a factor of the sum 49125+49124+……+492+49+149^{125} + 49^{124} + \ldots\ldots + 49^{2} + 49 + 149125+49124+……+492+49+1, is:
  1. (A)65
  2. (B)63
  3. (C)32
  4. (D)60

Correct answer: (B)

Step-by-step solution →
Q252·MathematicsMultiple correctJEE Advanced 2019
Let α\alphaα and β\betaβ be the roots of x2−x−1=0x^2 - x - 1 = 0x2−x−1=0, with α>β\alpha > \betaα>β. For all positive integers n, define an=αn−βnα−βa_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}an​=α−βαn−βn​, n≥1n \ge 1n≥1, b1=1b_1 = 1b1​=1 and bn=an−1+an+1b_n = a_{n-1} + a_{n+1}bn​=an−1​+an+1​, n≥2n \ge 2n≥2. Then which of the following options is/are correct?
  1. (A)∑n=1∞bn10n=889\sum_{n=1}^{\infty} \frac{b_n}{10^n} = \frac{8}{89}∑n=1∞​10nbn​​=898​
  2. (B)bn=αn+βnb_n = \alpha^n + \beta^nbn​=αn+βn for all n≥1n \ge 1n≥1
  3. (C)a1+a2+a3+.....+an=an+2−1a_1 + a_2 + a_3 + ..... + a_n = a_{n+2} - 1a1​+a2​+a3​+.....+an​=an+2​−1 for all n≥1n \ge 1n≥1
  4. (D)∑n=1∞an10n=1089\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{10}{89}∑n=1∞​10nan​​=8910​

Correct answer: (B), (C), (D)

Step-by-step solution →
Q253·MathematicsNumericalJEE Advanced 2019
Let AP(a; d) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d>0d > 0d>0. If AP(1;3)∩AP(2;5)∩AP(3;7)=AP(a;d)AP(1; 3) \cap AP(2; 5) \cap AP(3; 7) = AP(a; d)AP(1;3)∩AP(2;5)∩AP(3;7)=AP(a;d) then a+da + da+d equals ____

Correct answer: 157.00

Step-by-step solution →
Q254·MathematicsSingle correctJEE Main 2019
Let Sn_{n}n​ denote the sum of the first n terms of an A.P.. If S4_{4}4​ = 16 and S6_{6}6​ = −48-48−48, then S10_{10}10​ is equal to :
  1. (A)−410-410−410
  2. (B)−260-260−260
  3. (C)−320-320−320
  4. (D)−380-380−380

Correct answer: (C)

Step-by-step solution →
Q255·MathematicsSingle correctJEE Main 2019
If a1,a2,a3,……a_{1}, a_{2}, a_{3}, \ldots\ldotsa1​,a2​,a3​,…… are in A.P. such that a1+a7+a16=40a_{1} + a_{7} + a_{16} = 40a1​+a7​+a16​=40, then the sum of the first 15 terms of this A.P. is:
  1. (A)200
  2. (B)280
  3. (C)150
  4. (D)120

Correct answer: (A)

Step-by-step solution →
Q256·MathematicsSingle correctJEE Main 2019
If α\alphaα and β\betaβ are the roots of the equation 375x2−25x−2=0375x^{2}-25x-2=0375x2−25x−2=0, then lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\lim_{n \to \infty}\sum_{r=1}^{n}\alpha^{r} + \lim_{n \to \infty}\sum_{r=1}^{n}\beta^{r}limn→∞​∑r=1n​αr+limn→∞​∑r=1n​βr is equal to :
  1. (A)112\frac{1}{12}121​
  2. (B)29358\frac{29}{358}35829​
  3. (C)7116\frac{7}{116}1167​
  4. (D)21346\frac{21}{346}34621​

Correct answer: (A)

Step-by-step solution →
Q257·MathematicsSingle correctJEE Main 2019
The sum 1+13+231+2+13+23+331+2+3+…+13+23+33+…+1531+2+3+…+15−12(1+2+3+…+15)1 + \dfrac{1^{3} + 2^{3}}{1 + 2} + \dfrac{1^{3} + 2^{3} + 3^{3}}{1 + 2 + 3} + \ldots + \dfrac{1^{3} + 2^{3} + 3^{3} + \ldots + 15^{3}}{1 + 2 + 3 + \ldots + 15} - \dfrac{1}{2}\left(1 + 2 + 3 + \ldots + 15\right)1+1+213+23​+1+2+313+23+33​+…+1+2+3+…+1513+23+33+…+153​−21​(1+2+3+…+15) is equal to
  1. (A)620
  2. (B)1860
  3. (C)1240
  4. (D)660

Correct answer: (A)

Step-by-step solution →
Q258·MathematicsSingle correctJEE Main 2019
Let a1a_1a1​, a2a_2a2​, a3a_3a3​, …… be and A.P with a6a_6a6​ = 2. Then the common difference of this A.P., which maximizes the product a1a4a5a_1a_4a_5a1​a4​a5​ is
  1. (A)32\dfrac{3}{2}23​
  2. (B)85\dfrac{8}{5}58​
  3. (C)23\dfrac{2}{3}32​
  4. (D)65\dfrac{6}{5}56​

Correct answer: (B)

Step-by-step solution →
Q259·MathematicsSingle correctJEE Main 2019
The sum 3×112+5×(13+23)12+22+7×(13+23+33)12+22+32+........\frac{3 \times 1}{1^{2}} + \frac{5 \times (1^{3}+2^{3})}{1^{2}+2^{2}} + \frac{7 \times (1^{3}+2^{3}+3^{3})}{1^{2}+2^{2}+3^{2}} + ........123×1​+12+225×(13+23)​+12+22+327×(13+23+33)​+........ upto 10th^{th}th term, is
  1. (A)620
  2. (B)660
  3. (C)680
  4. (D)600

Correct answer: (B)

Step-by-step solution →
Q260·MathematicsSingle correctJEE Main 2019
Let a, b and c be in G.P with common ratio r, where a≠0a \neq 0a=0 and 0<r≤120 < r \leq \dfrac{1}{2}0<r≤21​. If 3a, 7b and 15c are the first three terms of an A.P., then the 4th4^{\text{th}}4th term of this A.P is
  1. (A)23a\dfrac{2}{3}a32​a
  2. (B)73a\dfrac{7}{3}a37​a
  3. (C)5a5a5a
  4. (D)aaa

Correct answer: (D)

Step-by-step solution →
Q261·MathematicsSingle correctJEE Main 2019
If a1_{1}1​, a2_{2}2​, a3_{3}3​ ........... an_{n}n​ are in A.P and a1_{1}1​ + a4_{4}4​ + a7_{7}7​ + ............... + a16_{16}16​ = 114, then a1_{1}1​ + a6_{6}6​ + a11_{11}11​ + a16_{16}16​ is equal to
  1. (A)76
  2. (B)64
  3. (C)98
  4. (D)38

Correct answer: (A)

Step-by-step solution →
Q262·MathematicsSingle correctJEE Main 2019
Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is
  1. (A)190
  2. (B)262
  3. (C)225
  4. (D)157

Correct answer: (A)

Step-by-step solution →
Q263·MathematicsSingle correctJEE Main 2019
The sum of the series 1+2×3+3×5+4×7+....1+2\times3+3\times5+4\times7+....1+2×3+3×5+4×7+.... upto 11th11^{th}11th term is
  1. (A)915
  2. (B)946
  3. (C)945
  4. (D)916

Correct answer: (B)

Step-by-step solution →
Q264·MathematicsSingle correctJEE Main 2019
Let ∑k=110f(a+k)=16(210−1)\displaystyle\sum_{k=1}^{10} f(a+k)=16(2^{10}-1)k=1∑10​f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)f(y)f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) for all natural numbers x, y and f(1)=2f(1)=2f(1)=2. Then the natural number 'a' is:
  1. (A)4
  2. (B)16
  3. (C)2
  4. (D)3

Correct answer: (D)

Step-by-step solution →
Q265·MathematicsSingle correctJEE Main 2019
If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th^{th}th tern is
  1. (A)−25-25−25
  2. (B)252525
  3. (C)−36-36−36
  4. (D)−35-35−35

Correct answer: (A)

Step-by-step solution →
Q266·MathematicsSingle correctJEE Main 2019
Let the sum of the first n terms of a non-constant A.P., a1,a2,a3,......a_1, a_2, a_3, ......a1​,a2​,a3​,...... be 50n+n(n−7)2A50n + \dfrac{n(n-7)}{2}A50n+2n(n−7)​A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d,a50)(d, a_{50})(d,a50​) is equal to:
  1. (A)(A,50+46A)(A, 50+46A)(A,50+46A)
  2. (B)(A,50+45A)(A, 50+45A)(A,50+45A)
  3. (C)(50,50+45A)(50, 50+45A)(50,50+45A)
  4. (D)(50,50+46A)(50, 50+46A)(50,50+46A)

Correct answer: (A)

Step-by-step solution →
Q267·MathematicsSingle correctJEE Main 2019
Then sum ∑k=120k12k\sum_{k=1}^{20}k\frac{1}{2^{k}}∑k=120​k2k1​ is equal to:
  1. (A)2−112192-\frac{11}{2^{19}}2−21911​
  2. (B)1−112201-\frac{11}{2^{20}}1−22011​
  3. (C)2−212202-\frac{21}{2^{20}}2−22021​
  4. (D)2−32172-\frac{3}{2^{17}}2−2173​

Correct answer: (A)

Step-by-step solution →
Q268·MathematicsSingle correctJEE Main 2019
The sum of all natural numbers 'n' such that 100<n<200100<n<200100<n<200 and H.C. F (91, n) > 1 is:
  1. (A)3221
  2. (B)3303
  3. (C)3203
  4. (D)3121

Correct answer: (D)

Step-by-step solution →
Q269·MathematicsSingle correctJEE Main 2019
Let Sk=1+2+3+.....+kkS_{k}=\frac{1+2+3+.....+k}{k}Sk​=k1+2+3+.....+k​. If S12+S22+.....+S102=512AS_{1}^{2}+S_{2}^{2}+.....+S_{10}^{2}=\frac{5}{12}AS12​+S22​+.....+S102​=125​A, then A equal to:
  1. (A)283
  2. (B)301
  3. (C)303
  4. (D)156

Correct answer: (C)

Step-by-step solution →
Q270·MathematicsSingle correctJEE Main 2019
The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is:
  1. (A)36
  2. (B)32
  3. (C)24
  4. (D)28

Correct answer: (D)

Step-by-step solution →
Q271·MathematicsSingle correctJEE Main 2019
If the sum of the first 15 terms of the series (34)3+(112)3+(214)3+33+(334)3+.....\left(\frac{3}{4}\right)^{3} + \left(1\frac{1}{2}\right)^{3} + \left(2\frac{1}{4}\right)^{3} + 3^{3} + \left(3\frac{3}{4}\right)^{3} + .....(43​)3+(121​)3+(241​)3+33+(343​)3+..... is equal to 225 k, then k is equal to :
  1. (A)108
  2. (B)27
  3. (C)54
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q272·MathematicsSingle correctJEE Main 2019
Let a1,a2,…,a10a_{1}, a_{2}, \ldots, a_{10}a1​,a2​,…,a10​ be a G.P. If a3a1=25\frac{a_{3}}{a_{1}}=25a1​a3​​=25, then a9a5\frac{a_{9}}{a_{5}}a5​a9​​ equals:
  1. (A)545^{4}54
  2. (B)4(52)4(5^{2})4(52)
  3. (C)535^{3}53
  4. (D)2(52)2(5^{2})2(52)

Correct answer: (A)

Step-by-step solution →
Q273·MathematicsSingle correctJEE Main 2019
If 19th19^{th}19th terms of non – zero A.P. is zero, then its (49th49^{th}49th term) : (29th29^{th}29th term) is:
  1. (A)4:1
  2. (B)1:3
  3. (C)3:1
  4. (D)2:1

Correct answer: (C)

Step-by-step solution →
Q274·MathematicsSingle correctJEE Main 2019
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 2719\frac{27}{19}1927​. Then the common ratio of this series is:
  1. (A)13\frac{1}{3}31​
  2. (B)23\frac{2}{3}32​
  3. (C)29\frac{2}{9}92​
  4. (D)49\frac{4}{9}94​

Correct answer: (B)

Step-by-step solution →
Q275·MathematicsSingle correctJEE Main 2019
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is:
  1. (A)1256
  2. (B)1465
  3. (C)1365
  4. (D)1356

Correct answer: (D)

Step-by-step solution →
Q276·MathematicsSingle correctJEE Main 2019
If 5, 5r, 5r2^{2}2 are the lengths of the sides of a triangle, then r cannot be equal to:
  1. (A)34\frac{3}{4}43​
  2. (B)54\frac{5}{4}45​
  3. (C)74\frac{7}{4}47​
  4. (D)32\frac{3}{2}23​

Correct answer: (C)

Step-by-step solution →
Q277·MathematicsSingle correctJEE Main 2019
The sum of the following series 1+6+9(12+22+32)7+12(12+22+32+42)9+15(12+22+....+52)11+...1+6+\frac{9(1^{2}+2^{2}+3^{2})}{7}+\frac{12(1^{2}+2^{2}+3^{2}+4^{2})}{9}+\frac{15(1^{2}+2^{2}+....+5^{2})}{11}+...1+6+79(12+22+32)​+912(12+22+32+42)​+1115(12+22+....+52)​+... up to 15 terms, is:
  1. (A)7820
  2. (B)7830
  3. (C)7520
  4. (D)7510

Correct answer: (A)

Step-by-step solution →
Q278·MathematicsSingle correctJEE Main 2019
Let a, b and c be the 7th7^{th}7th, 11th11^{th}11th and 13th13^{th}13th terms respectively of a non −-− constant A.P. If these are also the three consecutive terms of a G.P. then ac\frac{a}{c}ca​ is equal to:
  1. (A)12\frac{1}{2}21​
  2. (B)4
  3. (C)2
  4. (D)713\frac{7}{13}137​

Correct answer: (B)

Step-by-step solution →
Q279·MathematicsSingle correctJEE Main 2019
If a, b and c be three distinct numbers in G.P. and a+b+c=xba + b + c = xba+b+c=xb then x can not be
  1. (A)−2-2−2
  2. (B)−3-3−3
  3. (C)444
  4. (D)222

Correct answer: (D)

Step-by-step solution →
Q280·MathematicsNumericalJEE Advanced 2018
Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, ….. , and Y be the set consisting of the first 2018 terms of arithmetic progression 9, 16, 23, ….. . Then, the number of elements in the set X∪YX \cup YX∪Y is ______ .

Correct answer: 3748

Step-by-step solution →
Q281·MathematicsIntegerJEE Advanced 2017
The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side ?

Correct answer: 6

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Q282·MathematicsSingle correctJEE Advanced 2017
PARAGRAPH 2 Let ppp, qqq be integers and let α\alphaα, β\betaβ be the roots of the equation x2−x−1=0x^{2} - x - 1 = 0x2−x−1=0, where α≠β\alpha \ne \betaα=β. For n=0,1,2,…n = 0, 1, 2, \dotsn=0,1,2,…, let an=pαn+qβna_{n} = p\alpha^{n} + q\beta^{n}an​=pαn+qβn. FACT: If aaa and bbb are rational numbers and a+b5=0a + b\sqrt{5} = 0a+b5​=0, then a=0=ba = 0 = ba=0=b. a12=a_{12} =a12​=
  1. (A)a11−a10a_{11} - a_{10}a11​−a10​
  2. (B)a11+a10a_{11} + a_{10}a11​+a10​
  3. (C)2a11+a102a_{11} + a_{10}2a11​+a10​
  4. (D)a11+2a10a_{11} + 2a_{10}a11​+2a10​

Correct answer: (B)

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Q283·MathematicsSingle correctJEE Advanced 2016
Let bi>1b_{i} > 1bi​>1 for i=1,2,…,101i = 1, 2, \ldots, 101i=1,2,…,101. Suppose log⁡eb1,log⁡eb2,…,log⁡eb101\log_{e} b_{1}, \log_{e} b_{2}, \ldots, \log_{e} b_{101}loge​b1​,loge​b2​,…,loge​b101​ are in Arithmetic Progression (A. P.) with the common difference log⁡e2\log_{e} 2loge​2. Suppose a1,a2,…,a101a_{1}, a_{2}, \ldots, a_{101}a1​,a2​,…,a101​ are in A.P. such that a1=b1a_{1} = b_{1}a1​=b1​ and a51=b51a_{51} = b_{51}a51​=b51​. If t=b1+b2+…+b51t = b_{1} + b_{2} + \ldots + b_{51}t=b1​+b2​+…+b51​ and s=a1+a2+…+a51s = a_{1} + a_{2} + \ldots + a_{51}s=a1​+a2​+…+a51​, then
  1. (A)s>ts > ts>t and a101>b101a_{101} > b_{101}a101​>b101​
  2. (B)s>ts > ts>t and a101<b101a_{101} < b_{101}a101​<b101​
  3. (C)s<ts < ts<t and a101>b101a_{101} > b_{101}a101​>b101​
  4. (D)s<ts < ts<t and a101<b101a_{101} < b_{101}a101​<b101​

Correct answer: (B)

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Q284·MathematicsIntegerJEE Advanced 2015
Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

Correct answer: 9

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Q285·MathematicsIntegerJEE Advanced 2014
Let a,b,ca, b, ca,b,c be positive integers such that ba\frac{b}{a}ab​ is an integer. If a,b,ca, b, ca,b,c are in geometric progression and the arithmetic mean of a,b,ca, b, ca,b,c is b+2b + 2b+2, then the value of a2+a−14a+1\frac{a^2 + a - 14}{a+1}a+1a2+a−14​ is __________

Correct answer: 4

Step-by-step solution →
Q286·MathematicsIntegerJEE Advanced 2013
A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k−20=k-20=k−20= ________

Correct answer: 5

Step-by-step solution →
Q287·MathematicsMultiple correctJEE Advanced 2013
Let Sn=∑k=14n(−1)k(k+1)2k2S_n=\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2Sn​=∑k=14n​(−1)2k(k+1)​k2. Then SnS_nSn​ can take value(s)
  1. (A)105610561056
  2. (B)108810881088
  3. (C)112011201120
  4. (D)133213321332

Correct answer: (A), (D)

Step-by-step solution →

Sequence and Series — frequently asked

How many questions from Sequence and Series appear in JEE?

Sequence and Series has appeared in 164 of the last 186 JEE Main and JEE Advanced papers — about 88% of them — contributing 287 questions in total across those papers.

Is Sequence and Series an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 88% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Sequence and Series questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226
  • Permutations and Combinations 220

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