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🔢 Quant

Sequences & Series

Arithmetic, geometric and special series.

7%
of Quant

Why This Topic Matters

Total PYQs📊
22
of 1002 · 2021–2025
Years featured📅
5/5
of recent CAT years
% of Quant📈
~7%
of section questions
Est. hours⏱️
~8h
to master
~2/22
2021
~2/22
2022
2/22
2023
~2/22
2024
~2/22
2025
🎯PYQ Evidence

CAT 2021–2025: ~1.5 per slot (2021: 1.3 · 2022: 1.3 · 2023: 2.0 · 2024: 1.3 · 2025: 1.3). Progressions appeared every year (1–2 per slot) — AP/GP sums, common terms of two APs, and telescoping are the staples.

Sequences & Series

CAT favours the two classic progressions — arithmetic (constant difference) and geometric (constant ratio) — plus a few special sums.

Arithmetic progression (AP)

nn-th term   tn=a+(n1)d  \;t_n=a+(n-1)d\; and sum

Sn=n2[2a+(n1)d]=n2(first+last).S_n=\frac n2\big[\,2a+(n-1)d\,\big]=\frac n2\,(\text{first}+\text{last}).

Geometric progression (GP)

nn-th term   tn=arn1  \;t_n=a\,r^{\,n-1}\; and sum

Sn=arn1r1 (r1);S=a1r for r<1.S_n=a\,\frac{r^n-1}{r-1}\ (r\ne1);\qquad S_\infty=\frac{a}{1-r}\ \text{for}\ |r|<1.

A worked example

Find the sum of the first 20 terms of 3,7,11,15,3,\,7,\,11,\,15,\dots

This is an AP with a=3a=3, d=4d=4, n=20n=20:

S20=202[2(3)+19(4)]=10(6+76)=1082=820.S_{20}=\frac{20}{2}\big[\,2(3)+19(4)\,\big]=10\,(6+76)=10\cdot82=\mathbf{820}.

Special sums worth memorising

k=1nk=n(n+1)2,k2=n(n+1)(2n+1)6,k3=(n(n+1)2)2.\sum_{k=1}^{n}k=\frac{n(n+1)}2,\quad \sum k^2=\frac{n(n+1)(2n+1)}6,\quad \sum k^3=\left(\frac{n(n+1)}2\right)^2.

🎯PYQ Evidence
Find the hidden structure — a block, a difference of sums, or a clean doubling. : the value of floor(√n) stays constant between consecutive perfect squares, with exactly 2k+1 values of n giving k, so you add a handful of blocks instead of 50 terms and reach 217. : recover the nth term as a_n = S_n − S_{n−1} = 4n − 1, then "divisible by 9" is just a remainder condition mod 9, giving the smallest n = 7. : the messy rule a_n = 2a_{n−1} + 3 becomes exact doubling once you add 3 — a_n + 3 = 2(a_{n−1} + 3) — so a_n + 3 = 5·2^(n−1), and crossing one million needs n = 19. Reshape the sequence into something that adds or doubles cleanly and the bound falls out.

Common traps

  • AP vs GP mix-up. Check whether terms differ by adding (AP) or multiplying (GP).
  • Counting terms. From term pp to term qq inclusive there are qp+1q-p+1 terms — off-by-one errors are common.
  • Infinite GP condition. SS_\infty exists only when r<1|r|<1.

Checklist

  • Identify AP (common difference) vs GP (common ratio)
  • Sn=n2(first+last)S_n=\frac n2(\text{first}+\text{last}) for any AP
  • Use a(rn1)r1\frac{a(r^n-1)}{r-1} for GP; a1r\frac{a}{1-r} for infinite
  • Keep the k, k2, k3\sum k,\ \sum k^2,\ \sum k^3 formulas ready

Sample Questions

31 practice questions

Easy

Suppose x1x_{1}, x2x_{2}, x3x_{3}, …, x13x_{13} are in arithmetic progression such that x6x_{6} = -6 and 2x4x_{4} + 2x8x_{8} = x7x_{7} + x11x_{11}. Then, x75x_{75} equals

Easy

For some positive and distinct real numbers a, b and c, if 1/(b\sqrt{b}+c\sqrt{c}) is the arithmetic mean of 1/(a\sqrt{a}+c\sqrt{c}) and 1/(a\sqrt{a}+b\sqrt{b}), then the relationship which will always hold true is

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CAT PYQ Spotlight

Actual CAT questions on this topic

CAT 2025 · Slot 1
Medium

In the set of consecutive odd numbers {1, 3, 5, ……, 57}, there is a number of k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k. Then, k equals.

CAT 2024 · Slot 1
TITAMedium

For any natural number n, let ana_{n} be the largest integer not exceeding n\sqrt{n}. Then the value of a1a_{1} + a2a_{2} + … + a50a_{50} is

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