Geometric Progressions & Infinite Series: JEE Main

Geometric progressions (GP) are a central topic in JEE Main Mathematics, contributing one to two questions per session alongside arithmetic progressions. The chapter connects to compound interest, limits of infinite series, and logarithms. Understanding GP deeply also builds intuition for exponential functions and convergence, which appears in calculus. This guide covers all the results and question types JEE Main uses.

GP Definition and Standard Results

A geometric progression is a sequence where each term is obtained by multiplying the previous term by a constant ratio r. General term: T_n = ar^(n−1), where a is the first term. Sum of n terms: S_n = a(1−rⁿ)/(1−r) for r ≠ 1; S_n = na for r = 1. Infinite GP (|r| < 1): S_∞ = a/(1−r). This infinite series result is one of the most versatile in JEE Main — it converts repeating decimals to fractions and appears in probability problems involving geometric distributions.

Standard GP facts: product of first and last term equals product of any equidistant pair from ends. Geometric mean GM of two numbers a and b: GM = √(ab). For n numbers in GP: product of all n terms = (first term × last term)^(n/2). Three numbers in GP: a/r, a, ar (geometric mean is the middle term). For the arithmetic progression complement, see our Sequences and Series Guide.

Sum of Special Series

Telescoping series: Σ(1/(n(n+1))) = Σ(1/n − 1/(n+1)) = 1 − 1/(N+1) = N/(N+1) for n from 1 to N. More generally: Σ(1/(n(n+k))) = (1/k)·Σ(1/n − 1/(n+k)). JEE Main uses partial fractions to convert the general term into a telescoping form. Sums of powers: Σn = n(n+1)/2; Σn² = n(n+1)(2n+1)/6; Σn³ = [n(n+1)/2]² = (Σn)². These three sums appear in definite integral limits as Riemann sums in calculus questions. Take a free mock test on sequences and series to practise GP and special series problems.

Arithmetic-Geometric Progression (AGP)

An AGP has terms of the form n·rⁿ⁻¹ or (an+b)·rⁿ⁻¹ — a combination of an arithmetic factor and a geometric factor. Sum of AGP: use the "multiply by r and subtract" technique. Example: S = 1 + 2x + 3x² + 4x³ + ... to n terms. Then xS = x + 2x² + 3x³ + ... Subtract: S(1−x) = 1 + x + x² + ... − nx^n = (1−xⁿ)/(1−x) − nxⁿ. Solve for S. For infinite AGP (|x| < 1): S_∞ = 1/(1−x)². JEE Main tests the S_∞ = 1/(1−x)² result explicitly and as part of larger problems.

Relationships Between AM and GM

For positive numbers: AM ≥ GM (Arithmetic Mean ≥ Geometric Mean), with equality when all numbers are equal. For two positive numbers a and b: (a+b)/2 ≥ √(ab). JEE Main uses AM-GM to find minimum values: if x > 0, find minimum of x + 1/x. By AM-GM: x + 1/x ≥ 2√(x·1/x) = 2, equality at x = 1. This minimum-value technique appears in optimization problems and in proving inequalities. The AM-GM inequality is also used in coordinate geometry (finding minimum distance problems).

GP and Logarithms

If a, b, c are in GP, then log a, log b, log c are in AP (since b = ar → log b = log a + log r, and c = ar² → log c = log a + 2log r). This connection between GP and AP via logarithms is tested in JEE Main: "If a, b, c are in GP and log a, log b, log c are in AP, prove that the common difference of the AP is log r where r is the common ratio of the GP." Conversely, if a, b, c are in AP, then 10^a, 10^b, 10^c are in GP. For logarithm properties used here, see our Logarithms Guide.

Exam Strategy

GP problems in JEE Main are usually solved in 2-3 steps: identify r and a (or the relation between them), write the sum formula, substitute. The most common trap is using r instead of 1/r when the series goes "backwards" or dealing with descending GP. For AGP, the "multiply by r and subtract" technique is mechanical — practise until it is automatic. For the complete sequences syllabus, pair with our Binomial Theorem Guide which uses series expansions. Upgrade for ₹149/month for 150+ GP and series problems with full solutions.