Sequences & Series67 questions13 PYQ

Sequences & Series — JEE Maths Practice Questions & Solutions

67 questions on Sequences & Series with full step-by-step solutions, including past-year (PYQ) problems. Free to practice.

mediumPYQ · JEE Main 2026
Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0, where p,q∈Zp, q \in \mathbb{Z}. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in G.P., then ∣p+q∣|p + q| equals:
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mediumPYQ · JEE Main 2026
If the sum of the first 10 terms of the series 11+14⋅4+21+24⋅4+31+34⋅4+⋯\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to:
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mediumPYQ · JEE Main 2026
Let AA be the set of first 101101 terms of an A.P. whose first term is 11 and common difference is 55, and let BB be the set of first 7171 terms of an A.P. whose first term is 99 and common difference is 77. Then the number of elements in A∩BA\cap B which are divisible by 33 is
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mediumPYQ · JEE Main 2026
Let A1,A2,…,A39A_1, A_2, \ldots, A_{39} be 39 arithmetic means between 59 and 159. Then the mean of A25A_{25}, A28A_{28}, A31A_{31} and A36A_{36} is equal to:
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mediumPYQ · JEE Main 2026
The sum 131+13+231+3+13+23+331+3+5+⋯\dfrac{1^3}{1}+\dfrac{1^3+2^3}{1+3}+\dfrac{1^3+2^3+3^3}{1+3+5}+\cdots up to 8 terms, is
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mediumPYQ · JEE Main 2026
Let a1,a2,a3,…a_1,a_2,a_3,\ldots be an A.P. and g1=a1,g2,g3,…g_1=a_1,g_2,g_3,\ldots be an increasing G.P. If a1=a2+g2=1a_1=a_2+g_2=1 and a3+g3=4a_3+g_3=4, then a10+g5a_{10}+g_5 is equal to
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mediumPYQ · JEE Main 2026
Let α=14+18+116+⋯∞\alpha=\dfrac14+\dfrac18+\dfrac1{16}+\cdots\infty and β=13+19+127+⋯∞\beta=\dfrac13+\dfrac19+\dfrac1{27}+\cdots\infty. Then the value of (0.2)log⁡5(α)+(0.04)log⁡5(β)(0.2)^{\log_{\sqrt5}(\alpha)}+(0.04)^{\log_5(\beta)} is
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mediumPYQ · JEE Main 2026
If ∑k=1nak=6n3\displaystyle\sum_{k=1}^{n}a_k=6n^3, then ∑k=16(ak+1−ak36)2\displaystyle\sum_{k=1}^{6}\left(\dfrac{a_{k+1}-a_k}{36}\right)^2 is equal to
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mediumPYQ · JEE Main 2026
The value of 13−23+33−⋯+1531^3-2^3+3^3-\cdots+15^3 is
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mediumPYQ · JEE Main 2026
The sum of the first ten terms of an A.P. is 160160 and the sum of the first two terms of a G.P. is 88. 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 the common difference of the A.P., then the sum of all possible values of the first term of the G.P. is
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easyPYQ · JEE Main 2026
∑n=110(528n(n+1)(n+2))\displaystyle\sum_{n=1}^{10}\left(\dfrac{528}{n(n+1)(n+2)}\right) is equal to
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easyPYQ · JEE Main 2026
Let the sum of the first nn terms of an A.P. be 3n2+5n3n^2+5n. Then the sum of squares of the first 1010 terms of the A.P. is
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easyPYQ · JEE Main 2026
The first term of an A.P. of 3030 non-negative terms is 103\dfrac{10}{3}. If the sum of this A.P. is the cube of its last term, then its common difference is
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expert
If
(1+3+5+⋯+a)+(1+3+5+⋯+b)=(1+3+5+⋯+c),(1 + 3 + 5 + \cdots + a) + (1 + 3 + 5 + \cdots + b) = (1 + 3 + 5 + \cdots + c),
such that (i) a+b+c=57a + b + c = 57 and (ii) a<10a < 10, then find the value of (2b−2c+a)(2b - 2c + a).
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hard
The value of ∑n=12008n4+2n3+3n2+2n+1n(n+1)\displaystyle\sum_{n=1}^{2008}\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n(n+1)} is
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hard
If positive numbers xx, yy, zz are in A.P., then the minimum value of
x+y2y−x+y+z2y−z\frac{x+y}{2y-x} + \frac{y+z}{2y-z}
is less than
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hard
Match Column-I with Column-II.
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hard
Let AiA_i (i=1,2,…,6)(i = 1, 2, \ldots, 6) represent 66 arithmetic means, GiG_i (i=1,2,…,6)(i = 1, 2, \ldots, 6) represent 66 geometric means and HiH_i (i=1,2,…,6)(i = 1, 2, \ldots, 6) represent 66 harmonic means between the two numbers 22 and 7272, and let GG represent the single geometric mean between 22 and 7272. Then the value of
(∏i=16Gi)(∑i=161Hi)∑i=16Ai\frac{\left(\displaystyle\prod_{i=1}^{6} G_i\right)\left(\displaystyle\sum_{i=1}^{6} \frac{1}{H_i}\right)} {\displaystyle\sum_{i=1}^{6} A_i}
is
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hard
If
12+2⋅22+32+2⋅42+52+2⋅62+⋯1^{2} + 2\cdot 2^{2} + 3^{2} + 2\cdot 4^{2} + 5^{2} + 2\cdot 6^{2} + \cdots
up to nn terms is equal to 15751575, then find the value of nn.
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hard
If
1+(1+2)(1+3)22(13+23)+(1+2+3)(1+3+5)32(13+23+33)+(1+2+3+4)(1+3+5+7)42(13+23+33+43)+⋯1 + \frac{(1+2)(1+3)}{2^{2}\left(1^{3}+2^{3}\right)} + \frac{(1+2+3)(1+3+5)}{3^{2}\left(1^{3}+2^{3}+3^{3}\right)} + \frac{(1+2+3+4)(1+3+5+7)}{4^{2}\left(1^{3}+2^{3}+3^{3}+4^{3}\right)} + \cdots
up to 1010 terms is equal to ab\dfrac{a}{b}, where aa and bb are relatively prime, then a−ba - b is equal to
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medium
Let a1,a2,…,a4001a_1, a_2, \ldots, a_{4001} be in A.P. such that 1a1a2+1a2a3+⋯+1a4000a4001=10\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10 and a2+a4000=50a_2+a_{4000}=50. Then ∣a1−a4001∣|a_1-a_{4001}| equals
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medium
If p+q⋅5xp−q⋅5x=q+r⋅5xq−r⋅5x=r+s⋅5xr−s⋅5x\frac{p+q\cdot 5^x}{p-q\cdot 5^x}=\frac{q+r\cdot 5^x}{q-r\cdot 5^x}=\frac{r+s\cdot 5^x}{r-s\cdot 5^x}, then p,q,r,sp, q, r, s are in
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medium
A sequence AnA_n is defined by A1=12A_1=\frac12 and An=2n−32nAn−1A_n=\frac{2n-3}{2n}A_{n-1} for n≥2n\ge 2. Then
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medium
The sequence {xn}\{x_n\} is defined by xk+1=xk2+xkx_{k+1}=x_k^2+x_k and x1=12x_1=\frac12. Then [1x1+1+1x2+1+⋯+1x100+1]\left[\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_{100}+1}\right], where [ ⋅ ][\,\cdot\,] is the greatest integer function, equals
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medium
If a,b,c,da, b, c, d are positive reals with abcd=1abcd=1, then the minimum value of (1+a)(1+b)(1+c)(1+d)(1+a)(1+b)(1+c)(1+d) is
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medium
Let a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 be five terms of a geometric sequence satisfying
0<a1<a2<a3<a4<a5<1000<a_1<a_2<a_3<a_4<a_5<100
where each term is an integer. Then the number of such geometric progressions is
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medium
If a1,a2,…,ana_1, a_2, \ldots, a_n are distinct odd natural numbers not divisible by any prime greater than 55, then
1a1+1a2+⋯+1an\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}
is less than
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medium
If a,b,c,da, b, c, d are in G.P., then (a2+b2+c2)(b2+c2+d2)(a^2+b^2+c^2)(b^2+c^2+d^2) equals
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medium
If ai>0a_i>0 for 1≤i≤51\le i\le 5 and a1a2a3a4a5=6a_1 a_2 a_3 a_4 a_5=6, then the least value of a1+2a2+3a3+4a4+5a5a_1+2a_2+3a_3+4a_4+5a_5 is
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medium
If a,b,ca, b, c are positive with a+b+c=6a+b+c=6, then the minimum value of (a+1b)2+(b+1c)2+(c+1a)2\left(a+\frac1b\right)^2+\left(b+\frac1c\right)^2+\left(c+\frac1a\right)^2 is
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medium
The sum of the series 419+44192+444193+⋯\frac{4}{19}+\frac{44}{19^2}+\frac{444}{19^3}+\cdots to infinity is
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medium
If a+b+c+d+e=8a+b+c+d+e=8 and a2+b2+c2+d2+e2=16a^2+b^2+c^2+d^2+e^2=16, where a,b,c,d,ea,b,c,d,e are non-negative reals and the range of ee is [ℓ,mn]\left[\ell,\frac{m}{n}\right] with gcd⁡(m,n)=1\gcd(m,n)=1, then ℓ+m+n\ell+m+n is
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medium
If A,G,HA, G, H are respectively the A.M., G.M. and H.M. between two positive numbers, and xA=yG=zHxA=yG=zH where x,y,zx, y, z are non-zero positive quantities, then x,y,zx, y, z are in
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medium
The 20082008th term of the sequence 1, 2,2,2, 3,3,3,3,3,3, 4,…1,\ 2,2,2,\ 3,3,3,3,3,3,\ 4,\ldots where nn occurs n(n+1)2\frac{n(n+1)}{2} times, equals
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medium
If H1,H2,…,H20H_1, H_2, \ldots, H_{20} are 2020 harmonic means between 22 and 33, then H1+2H1−2+H20+3H20−3=\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=
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medium
If 112+122+132+⋯=π26\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\frac{\pi^2}{6}, then 112+132+152+⋯\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots is equal to
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medium
In a sequence of (4n+1)(4n+1) terms, the first (2n+1)(2n+1) terms are in A.P. with common difference 22, and the last (2n+1)(2n+1) terms are in G.P. with common ratio 0.50.5. If the middle terms of the A.P. and the G.P. are equal, then the middle term of the sequence is
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medium
The sum of the series x1−x2+x21−x4+x41−x8+⋯\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots to infinity, for ∣x∣<1|x|<1, is
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medium
If nn arithmetic means are inserted between aa and 2b2b, and also between 2a2a and bb (a,b∈Ra,b\in\mathbb{R}), and the mmth means of the two sets are equal, then a:ba:b equals
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medium
If a,b,ca, b, c are in A.P., p,p′p, p' are the A.M. and G.M. between aa and bb, and q,q′q, q' are the A.M. and G.M. between bb and cc, then
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medium
Through the centroid of an equilateral triangle a line parallel to the base is drawn. On this line an arbitrary interior point PP is taken. Let hh be the distance of PP from the base, and h1,h2h_1, h_2 its distances from the other two sides. Then
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medium
If the common ratio of a geometric progression is positive, then the maximum value of the ratio of the second term to the sum of the first three terms equals
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medium
The minimum value of
12x+18y+xy,\frac{12}{x} + \frac{18}{y} + xy ,
where xx and yy are positive real numbers, is
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medium
The largest term of the sequence 1503,4524,9581,16692,…\frac{1}{503},\frac{4}{524},\frac{9}{581},\frac{16}{692},\ldots is
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medium
In a triangle ABCABC, sin⁡2A+sin⁡2B=sin⁡2C\sin^{2}A+\sin^{2}B=\sin^{2}C and sin⁡A, sin⁡B, sin⁡C\sin A,\ \sin B,\ \sin C are in A.P. in that order. If the inradius of the triangle is 22, then which of the following is/are correct?
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easy
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
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easy
Let a1,a2,…a_1, a_2, \ldots and b1,b2,…b_1, b_2, \ldots be arithmetic progressions with a1=25a_1=25, b1=75b_1=75 and a100+b100=100a_{100}+b_{100}=100. Then the sum of the first hundred terms of a1+b1, a2+b2, …a_1+b_1,\ a_2+b_2,\ \ldots is
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easy
If Sn=11⋅2⋅3+12⋅3⋅4+⋯+1n(n+1)(n+2)S_n=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{n(n+1)(n+2)}, then 8S∞8S_\infty is
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easy
If the sum to infinity of 1+4x+7x2+10x3+⋯1+4x+7x^2+10x^3+\cdots is 3516\frac{35}{16}, where ∣x∣<1|x|<1, then xx equals
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easy
If tr=12+22+⋯+r213+23+⋯+r3t_r=\frac{1^2+2^2+\cdots+r^2}{1^3+2^3+\cdots+r^3} and Sn=∑r=1n(−1)rtrS_n=\sum_{r=1}^{n}(-1)^r t_r, then lim⁡n→∞Sn\lim_{n\to\infty}S_n equals
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easy
If Sn=1⋅23!+2⋅224!+3⋅235!+⋯S_n=\frac{1\cdot2}{3!}+\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\cdots up to nn terms, then the sum to infinite terms is
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easy
The sum of the first nn terms (n>1n>1) of an A.P. is 153153 and the common difference is 22. If the first term is an integer, the number of possible values of nn is
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easy
If the lengths of the sides of a right triangle are in A.P., then the sines of the acute angles are
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easy
Let f(n),g(n)f(n), g(n) denote the sums of nn terms of the sequences 2,5,10,17,26,…2,5,10,17,26,\ldots and 2,6,12,20,…2,6,12,20,\ldots respectively. Then lim⁡n→∞f(n)g(n)\lim_{n\to\infty}\frac{f(n)}{g(n)} is
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easy
Consider α0,α1,α2,…\alpha_0, \alpha_1, \alpha_2, \ldots with α0=17.23\alpha_0=17.23, α1=33.23\alpha_1=33.23 and αr+2=αr+αr+12\alpha_{r+2}=\frac{\alpha_r+\alpha_{r+1}}{2} for all r≥0r\ge 0. The value of ∣α10−α9∣|\alpha_{10}-\alpha_9| is
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easy
If 1−13+15−17+19−111+⋯=π41-\frac13+\frac15-\frac17+\frac19-\frac{1}{11}+\cdots=\frac{\pi}{4}, then the value of 11⋅3+15⋅7+19⋅11+⋯\frac{1}{1\cdot3}+\frac{1}{5\cdot7}+\frac{1}{9\cdot11}+\cdots is
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easy
If a,b,ca, b, c are three positive numbers, then the minimum value of ab(a+b)+bc(b+c)+ca(c+a)abc\frac{ab(a+b)+bc(b+c)+ca(c+a)}{abc} is
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easy
If the sum of nn terms of an A.P. is cn(n−1)cn(n-1) with c≠0c\ne 0, then the sum of the squares of these terms is
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easy
If a,b,ca, b, c are three distinct numbers such that a,b,ca, b, c are in A.P. and b−a, c−b, ab-a,\ c-b,\ a are in G.P., then a:b:ca:b:c is
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easy
If ak=1k(k+1)a_k=\frac{1}{k(k+1)} for k=1,2,…,nk=1,2,\ldots,n, then (∑k=1nak)2\left(\sum_{k=1}^{n}a_k\right)^2 is equal to
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easy
The value of the series is ∑k=1∞1k!(∑n=1k2n−1)\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}\left(\sum_{n=1}^{k}2^{n-1}\right)
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easy
If 1, log⁡93x+48, log⁡9(3x−83)1,\ \log_9\sqrt{3^x+48},\ \log_9\left(3^x-\frac83\right) are in A.P., then the value of xx is
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easy
If a,b,ca, b, c are in H.P., then ab+c, bc+a, ca+b\frac{a}{b+c},\ \frac{b}{c+a},\ \frac{c}{a+b} are in
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easy
The ratio of the sum of the first three terms of a G.P. to the sum of the first six terms is 64:9164:91. The common ratio of the G.P. is
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easy
The sum of the series 3+5+9+17+33+⋯3+5+9+17+33+\cdots to nn terms is
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easy
If one A.M. AA and two G.M.s pp and qq are inserted between two numbers aa and bb, then which is true?
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easy
If the fourth term of a G.P. is 33, then the product of the first seven terms is
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