A geometric sequence (or progression) multiplies by a fixed common ratio r between terms. From the first term a, the nth term is aₙ = a × r^(n−1), and the sum of the first n terms is Sₙ = a(1 − rⁿ)/(1 − r). When |r| < 1, an infinite series even has a finite sum, a/(1 − r).
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the geometric nth-term and series-sum formulas, recomputed in code.
The formulas
The nth term scales the first by r raised to (n − 1). The finite-sum formula comes from the algebra of a geometric series. When the ratio is between −1 and 1 the terms shrink toward zero, so even an endless sum settles on a/(1 − r); if |r| ≥ 1 the terms don't shrink and the infinite sum has no limit.
Worked examples
a = 2, r = 3, n = 5:
Doubling, a = 1, r = 2, n = 10:
Halving, a = 100, r = 0.5, n = 4:
So tripling reaches a 5th term of 162 summing to 242, doubling reaches 512 summing to 1023, and halving from 100 gives 12.5 summing to 187.5.
Frequently Asked Questions
Each term multiplies the previous by a constant ratio r. 2, 6, 18, 54 has r = 3.
aₙ = a × r^(n−1). a = 2, r = 3 → a₅ = 2 × 81 = 162.
Sₙ = a(1 − rⁿ)/(1 − r). For a=2,r=3,n=5: 2(1−243)/(−2) = 242. If r = 1, sum = n×a.
Yes. r = 0.5 shrinks (100, 50, 25…); negative r alternates signs (1, −2, 4…).
If |r| < 1, S∞ = a/(1−r). a=1, r=0.5 → 2. If |r| ≥ 1, it diverges.