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Sequences & Series revision notes

Every idea in the Sequences & Series chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 26 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Arithmetic progression

nth term of an AP
18× asked, last 2026
an=a+(n−1)d,ap−aq=(p−q) d,kth from the end=l−(k−1) da_n=a+(n-1)d,\qquad a_p-a_q=(p-q)\,d,\qquad k\text{th from the end}=l-(k-1)\,d
dd = any term minus the one before; it may be negative or a fraction. a,b,ca,b,c in AP   ⟺  2b=a+c\iff 2b=a+c.
  • Not multipliers 1,2,3,…1,2,3,\dots for log⁡a1/2x+log⁡a1/3x+⋯\log_{a^{1/2}}x+\log_{a^{1/3}}x+\cdots. Since log⁡a1/kx=klog⁡ax\log_{a^{1/k}}x=k\log_a x, the first term already carries 2. Twenty terms give (2+3+⋯+21)log⁡ax=230log⁡ax(2+3+\dots+21)\log_a x=230\log_a x.
Sum of n terms
23× asked, last 2026
Sn=n2[2a+(n−1)d]=n2 (a+l)S_n=\frac n2\left[2a+(n-1)d\right]=\frac n2\,(a+l)
ll = nnth term. For odd nn, Sn=n×S_n=n\times(middle term): S7=7a4S_7=7a_4, S21=21a11S_{21}=21a_{11}.
258111417a+l=19a+l=19
The AP 2,5,…,172,5,\dots,17 (blue) and the same AP reversed (orange) make nn columns of height a+la+l: 2Sn=n(a+l)2S_n=n(a+l)
  • Two APs with Sn:Sn′S_n:S_n' given as a function of nn: the ratio of mmth terms is that function at n=2m−1n=2m-1. For one AP, Sm:Sn=m2:n2S_m:S_n=m^2:n^2 forces a=d/2a=d/2 and am:an=(2m−1):(2n−1)a_m:a_n=(2m-1):(2n-1).
  • Not 21 terms in S30−S10S_{30}-S_{10}. It is terms 11 to 30, only 20 terms: Sm−Sn=(m−n)(an+1+am)/2S_m-S_n=(m-n)(a_{n+1}+a_m)/2. 'The next 15 terms' after the first 25 is S40−S25S_{40}-S_{25}.
  • Quick method. Avoid finding aa and dd when you can: keep 2a+(n−1)d2a+(n-1)d as one block and match brackets. S3n=3S2nS_{3n}=3S_{2n} gives 2a=−(n−1)d2a=-(n-1)d, so S4n/S2n=6S_{4n}/S_{2n}=6.
Term from the sum formula
5× asked, last 2026
an=Sn−Sn−1 (n≥2),a1=S1,Sn=An2+Bn  ⟺  AP with d=2Aa_n=S_n-S_{n-1}\ (n\ge2),\qquad a_1=S_1,\qquad S_n=An^2+Bn\iff\text{AP with } d=2A
A constant term in SnS_n spoils only a1a_1. A cubic SnS_n gives a quadratic ana_n, which is not an AP.
Terms in AP: symmetric choice
5× asked, last 2026
a−d, a, a+d,a−3d, a−d, a+d, a+3d,a−2d,…,a+2da-d,\ a,\ a+d,\qquad a-3d,\ a-d,\ a+d,\ a+3d,\qquad a-2d,\dots,a+2d
The sum fixes aa at once; a product becomes a(a2−d2)a(a^2-d^2). Test every root for dd against all the data.
  • Not dd as the common difference when four terms are written a±da\pm d, a±3da\pm3d. Neighbours differ by 2d2d. Double it before you report the difference or build the next term.
Properties of an AP
8× asked, last 2025
ak+an+1−k=a1+an,Seven−Sodd=n2 d  (n even)a_k+a_{n+1-k}=a_1+a_n,\qquad S_{\text{even}}-S_{\text{odd}}=\frac n2\,d\ \ (n\text{ even})
Adding or multiplying by a constant keeps an AP; termwise, two APs add to an AP but multiply to one only if one is constant.
  • Not n⋅dn\cdot d for (even-place sum) − (odd-place sum). 2k2k terms form kk pairs, each differing by dd, so the gap is kdkd. Sums 120 and 72 with d=1d=1 mean 48 pairs, so n=96n=96.
Inserting n arithmetic means
4× asked, last 2026
d=b−an+1,Ak=a+k b−an+1,A1+⋯+An=n(a+b)2d=\frac{b-a}{n+1},\qquad A_k=a+k\,\frac{b-a}{n+1},\qquad A_1+\dots+A_n=\frac{n(a+b)}{2}
nn means between aa and bb: n+2n+2 terms and n+1n+1 gaps. The kkth mean is the (k+1)(k+1)th term.
  • Not d=(b−a)/nd=(b-a)/n. 39 means between 59 and 159 make 40 gaps, so d=2.5d=2.5. The mean of A25,A28,A31,A36A_{25},A_{28},A_{31},A_{36} is the term at the mean position, A30=59+75=134A_{30}=59+75=134.
Common terms of two APs
9× asked, last 2026
d=lcm⁡(d1,d2),last common term≤min⁡(l1,l2)d=\operatorname{lcm}(d_1,d_2),\qquad \text{last common term}\le\min(l_1,l_2)
Start at the first number in both lists. n(A∪B)=n(A)+n(B)−n(A∩B)n(A\cup B)=n(A)+n(B)-n(A\cap B).
4,9,…,1244,9,\dots,1243,6,…,1113,6,\dots,111999114: past 111
Common terms (red) step by lcm(5,3)=15\text{lcm}(5,3)=15 from 9 and stop at 99: 7 terms
  • Not the larger last term as the cap. A common term must lie in both lists. 4,9,…,1244,9,\dots,124 and 3,6,…,1113,6,\dots,111 share 9,24,…,999,24,\dots,99: 7 terms, since 114 is beyond 111.
Counting terms and conditional sums
10× asked, last 2026
(n−1) d=l−a,∑multiples of k ≤N=k m(m+1)2,  m=⌊Nk⌋(n-1)\,d=l-a,\qquad \sum_{\text{multiples of }k\,\le N}=k\,\frac{m(m+1)}{2},\ \ m=\left\lfloor\frac Nk\right\rfloor
Integer dd, fixed ends: n−1n-1 divides l−al-a. Unwanted terms: whole sum minus their own AP (inclusion–exclusion).
  • Not mm terms from TmT_m to T2mT_{2m}. From TpT_p to TqT_q there are q−p+1q-p+1 terms, so T20+⋯+T40T_{20}+\dots+T_{40} has 21 terms.
  • Quick method. Two jumps alternating, as in 3+4+8+9+13+14+⋯3+4+8+9+13+14+\cdots: add in pairs. The pair sums 7,17,27,…7,17,27,\dots form an AP with half as many terms; 40 terms give 10(14+190)=204010(14+190)=2040.

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Practise Sequences & Series

18 more concepts in 6 topics

Geometric progression, Means and harmonic progression, Arithmetico-geometric series, Special sums and method of differences and Telescoping and recursive sequences, plus ideas from exponential and log series. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Geometric progression

  • nth term and sum of a GP20×
  • Terms in GP: symmetric choice6×
  • Products of GP terms11×
  • Logs of a GP form an AP7×
  • Infinite GP15×
  • AP and GP conditions together15×

Means and harmonic progression

  • AM, GM and HM5×
  • AM–GM for minimum and maximum6×
  • Harmonic progression6×

Arithmetico-geometric series

  • AGP: the S − rS method19×

Special sums and method of differences

  • Sums of powers of natural numbers18×
  • Alternating and grouped sums8×
  • Method of differences10×

Telescoping and recursive sequences

  • Telescoping sums8×
  • Products in AP in the denominator17×
  • Quartic denominators8×
  • Root forms: rationalise2×
  • Linear recurrences and power sums16×
  • Also remember2

Also asked here: exponential and log series

  • Key points2

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