PYQ Vault

JEE Mains Maths · Sequences and Series

Infinite Geometric Series

The sum to infinity of a GP with ratio between −1 and 1, the series of its squares or alternate terms, and infinite GPs hidden inside exponents, logarithms, double sums and repeated figures.

Why this matters

Eighteen PYQs, split evenly. Half give an infinite GP outright, often with the sum of its squares. The other half hide one: in an exponent like cos²x + cos⁴x + …, in a chain of logarithms, or in shapes shrinking step by step. Two ideas cover the page.

Concept 1 of 2: Sum to infinity, and the series of squares

When ∣r∣<1|r|<1, the terms shrink to zero and S∞=a1−rS_\infty=\frac{a}{1-r}. The squares of the terms form another infinite GP, a2,a2r2,…a^2,a^2r^2,\dots, with sum a21−r2\frac{a^2}{1-r^2}. Dividing that by S∞S_\infty cancels one aa and leaves a1+r\frac{a}{1+r}. Alternate terms form a GP with ratio r2r^2.

Definition

  • S∞=a1−rS_\infty=\frac a{1-r}, only for ∣r∣<1|r|<1.
  • Squares: a21−r2\frac{a^2}{1-r^2}. Cubes: a31−r3\frac{a^3}{1-r^3}.
  • sum of squaresS∞=a1+r\frac{\text{sum of squares}}{S_\infty}=\frac a{1+r}.
  • Alternate terms from arkar^k: ark1−r2\frac{ar^k}{1-r^2}.

Sum to infinity

S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\quad |r|<1

Worked example

An infinite GP has sum 6, and the sum of the squares of its terms is 12. Find aa and rr.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 18 · Q77Moderate

Example 1 · Sequences and Series · Infinite Geometric Series

If the sum of an infinite GP a,ar,ar2,ar3,…a,ar,ar^{2},ar^{3},\ldots is 15 and the sum of the squares of its each term is 150 , then the sum of ar2,ar4,ar6,…ar^{2},ar^{4},ar^{6},\ldots is:

Check |r| < 1

A quadratic in rr may give a root with ∣r∣≥1|r|\ge1. The sum to infinity does not exist there, so that root is rejected.

Concept 2 of 2: Infinite GPs inside exponents, logarithms and figures

An exponent cos⁡2x+cos⁡4x+…\cos^2x+\cos^4x+\dots is an infinite GP with ratio cos⁡2x\cos^2x, summing to cot⁡2x\cot^2x. A chain log⁡x+log⁡x1/3+log⁡x1/9+…\log x+\log x^{1/3}+\log x^{1/9}+\dots is log⁡x\log x times 1+13+19+⋯=321+\frac13+\frac19+\dots=\frac32. A bracket-by-bracket series such as (a+b)+(a2+ab+b2)+…(a+b)+(a^2+ab+b^2)+\dots often regroups into two GPs. Figures that shrink by the same factor each step, like triangles joined at midpoints, give a GP of areas or perimeters.

Definition

  • ∑k≥1cos⁡2kx=cot⁡2x\sum_{k\ge1}\cos^{2k}x=\cot^2x; ∑k≥1sin⁡2kx=tan⁡2x\sum_{k\ge1}\sin^{2k}x=\tan^2x.
  • ∑k≥0log⁡xck=log⁡x1−c\sum_{k\ge0}\log x^{c^k}=\frac{\log x}{1-c} for ∣c∣<1|c|<1.
  • Regroup: a sum over brackets can often be written as ∑ai\sum a^i times ∑bj\sum b^j, or as two separate GPs.
  • Midpoint triangles: each area is 14\frac14 of the last, each perimeter 12\frac12.

The geometric series

1+x+x2+⋯=11−x,∣x∣<11+x+x^2+\dots=\frac{1}{1-x},\quad |x|<1

Worked example

3sin⁡2x+sin⁡4x+…=33^{\sin^2x+\sin^4x+\dots}=3 with 0<x<π20<x<\frac\pi2. Find xx.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 1 · Q79Moderate

Example 2 · Sequences and Series · Infinite Geometric Series

If e(cos⁡2x+cos⁡4x+cos⁡6x+…∞)log⁡e2e^{\left( \cos^{2}x+\cos^{4}x+\cos^{6}x+ \ldots\infty \right)\log_{e}2} satisfies the equation t2−9t+8=0t^{2}- 9t+ 8 = 0, then the value of 2sin⁡xsin⁡x+3cos⁡x(0<x<π2)\frac{2\sin x}{\sin x+\sqrt{3}\cos x}\left( 0 <x<\frac{\pi}{2} \right) is :

The ratio must stay below 1

cos⁡2x+cos⁡4x+…\cos^2x+\cos^4x+\dots diverges where cos⁡2x=1\cos^2x=1. Check the stated range keeps the ratio strictly between −1 and 1.

Summary — formulas & gotchas at a glance

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Test yourself on Sequences and Series

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