Arithmetic, Geometric and Arithmetico-Geometric Progressions
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Direct answer
Three patterns power nearly every series question in JEE Mathematics — the arithmetic, geometric and arithmetico-geometric progressions. An AP with first term a and difference d has nth term a + (n − 1)d and sum n/2 × (2a + (n − 1)d); a GP with ratio r has nth term ar^(n − 1), finite sum a(r^n − 1)/(r − 1), and infinite sum a/(1 − r) valid only for |r| < 1. An AGP — a, (a + d)r, (a + 2d)r^2, ... — is tamed by writing S, multiplying by r, subtracting, and collapsing the survivor into a/(1 − r) + dr/(1 − r)^2, the derivation every student should perform once and then trust.
What you must remember
- AP toolkit: T_n = a + (n − 1)d; S_n = n/2(2a + (n − 1)d) = n(first term + last term)/2; three unknown terms in AP are best written a − d, a, a + d.
- GP toolkit: T_n = ar^(n − 1); Sn = a(r^n − 1)/(r − 1) for r ≠ 1; S∞ = a/(1 − r) only when |r| < 1 — the most-invoked validity condition in the chapter.
- AGP infinite sum: a/(1 − r) + dr/(1 − r)^2; for example 1 + 2x + 3x^2 + ... = 1/(1 − x)^2 for |x| < 1.
- Standard power sums: Σn = n(n + 1)/2, Σn^2 = n(n + 1)(2n + 1)/6, Σn^3 = [n(n + 1)/2]^2 — memorised as a triple.
- Means: AM ≥ GM for positives with G^2 = AM × HM; the equality case (all numbers equal) powers most minimum-value questions.
- Method of differences: when consecutive terms of a series differ in a recognisable pattern (often an AP), list rows shifted and add column-wise to telescope.
- Harmonic link: reciprocals of an HP form an AP; the harmonic mean of a and b is 2ab/(a + b).
Deriving the AGP sum honestly
Write S = a + (a + d)r + (a + 2d)r^2 + ... and multiply by r: rS = ar + (a + d)r^2 + (a + 2d)r^3 + ... Every term of rS except the first partial survivors aligns under a term of S one place to the right. Subtract: S(1 − r) = a + dr + dr^2 + dr^3 + ... = a + dr/(1 − r), where the geometric tail converges precisely because |r| < 1. Hence S = a/(1 − r) + dr/(1 − r)^2. Test it on 1 + 2x + 3x^2 + ... with a = 1, d = 1, r = x: the formula gives 1/(1 − x) + x/(1 − x)^2 = (1 − x + x)/(1 − x)^2 = 1/(1 − x)^2, matching the series obtained by differentiating the geometric sum. The derivation explains why the convergence condition is not decoration — without |r| < 1 the tail dr + dr^2 + ... never settles and S does not exist.
How JEE frames series questions
JEE Main keeps to direct machinery: a sum formula, an infinite GP, a standard power sum combination such as Σ(n^2 + n). JEE Advanced builds series from recurrences or asks for telescoping — 1/(1 × 2) + 1/(2 × 3) + ... + 1/(n(n + 1)) = n/(n + 1) — and mixes AM-GM equalities into "least value" problems. The recurring losses: applying S_∞ when |r| ≥ 1, forgetting that three numbers in GP should be written a/r, a, ar so their product is simply a^3, and assuming a pattern is arithmetic from two terms alone. Word problems about sums of salaries or ladder rungs almost always intend an AP; spotting that early converts a paragraph into one formula.
Frequently asked questions
When is the infinite GP sum valid?
Only for |r| < 1; otherwise the partial sums grow without bound and S_∞ does not exist.
What is 1 + 2x + 3x^2 + ... for |x| < 1?
1/(1 − x)^2, from the AGP formula with a = 1, d = 1, r = x.
What is the sum of the first n perfect squares?
n(n + 1)(2n + 1)/6.
How should three unknown numbers in GP be represented?
As a/r, a, ar, because the product becomes a^3 and the sum a(1 + r + 1/r) — both symmetric in r.
What relation connects the AM, GM and HM of two positive numbers?
G^2 = AM × HM, and AM ≥ GM ≥ HM with equality only when the two numbers are equal.