Geometric Sequence Calculator

Background.

A geometric sequence multiplies rather than adds. Each term is the one before it multiplied by the same fixed factor — the common ratio, written r. Start at 2 with r = 3 and you get 2, 6, 18, 54, 162, 486, 1458, 4374. Start at 3 with r = 0.5 and you get 3, 1.5, 0.75, 0.375, and terms that keep halving forever without ever reaching zero. That single difference from an arithmetic sequence, multiply instead of add, is what makes geometric sequences the mathematics behind compound interest, radioactive decay, population growth, the halving of a signal through successive attenuators, and the reason a folded sheet of paper gets thick so alarmingly fast.

This calculator solves the progression in all four directions. Give it the first term, the ratio and a position and it returns the term there. Give it two terms and their separation and it recovers the ratio. Give it a target value and it finds which position that value occupies, or tells you plainly that the value is not on the sequence at all. Every mode also reports the sum of the run, the explicit rule, a preview of the terms, and — the output that has no counterpart on the arithmetic page — whether the corresponding infinite series converges.

That convergence verdict is the most interesting thing a geometric sequence does, and it is the reason this page reports it as a sentence rather than a number. When the ratio sits strictly between −1 and 1, the terms shrink toward zero fast enough that adding infinitely many of them still lands on a finite total, and OpenStax gives that limit as S = a₁/(1 − r). When the ratio is 1 or larger in size, the terms never shrink and the partial sums run away; there is no finite answer to print, so printing one would be a lie. The classic demonstration is a₁ = 0.9 with r = 0.1: the infinite sum 0.9 + 0.09 + 0.009 + … equals 0.9/0.9 = exactly 1, which is the standard proof that 0.999… and 1 are the same number. Type those two values in and the page says so.

The page is 1-indexed: a₁ is the first term, so the n-th term is a₁ × r^(n − 1) and the exponent is one less than the position. It is worth knowing that the Encyclopedia of Mathematics writes the same progression 0-indexed, as aⱼ = a₀qʲ, which describes an identical sequence with every index shifted down by one. If your source uses that convention, its a₀ is this page's a₁.

One genuine ambiguity is worth flagging before you read a ratio the calculator solved for. If you know a₁ and a later term separated by an even number of steps — say the first term is 1 and the third term is 16 — then both r = 4 and r = −4 produce that third term, because squaring kills the sign. Two real sequences fit your data, 1, 4, 16 and 1, −4, 16. The page returns the positive root and says so here rather than pretending the answer is unique. When the separation is an odd number of steps the root is unique and no ambiguity arises. And when the ratio of the two terms is negative while the separation is even, no real ratio exists at all — an even power of a real number is never negative — so the calculator refuses rather than inventing one.

If your terms increase by a fixed amount rather than a fixed factor, you want the arithmetic sequence calculator. If you have a list of numbers and are not sure which family they belong to, the sequence calculator identifies it. And if you want a sum whose index starts somewhere other than 1, or a sum of squares or cubes, the sum of series calculator handles sigma notation directly.

What is geometric sequence calculator?

A geometric sequence — also called a geometric progression — is a sequence in which each term is the previous term multiplied by a fixed non-zero number. OpenStax College Algebra 2e §9.3 puts it as: 'Each term of a geometric sequence increases or decreases by a constant factor called the common ratio.' The Encyclopedia of Mathematics is explicit that the multiplier must be non-zero: 'a sequence of numbers each one of which is equal to the preceding one multiplied by a number q ≠ 0.' The n-th term is aₙ = a₁r^(n−1). The sum of the first n terms — a finite geometric series — is Sₙ = a₁(1 − rⁿ)/(1 − r) whenever r ≠ 1, and simply n × a₁ when r = 1. If the ratio satisfies −1 < r < 1, the infinite series converges to S = a₁/(1 − r); otherwise it diverges. Geometric sequences describe anything that changes by a constant proportion per step: compound interest, radioactive decay measured in half-lives, bacterial doubling, depreciation at a fixed percentage per year, or the amplitude of a bouncing ball losing the same fraction of its height on each bounce.

How to use this calculator.

1. 01Pick what you want to find — the nth term, the common ratio, the first term, or the number of terms.
2. 02Fill in the three fields that mode needs. The field being solved for is ignored, so leave whatever is in it.
3. 03Read the hero number, then the partial sum, and then the infinite-series line, which tells you whether adding terms forever converges and to what.
4. 04Check the sequence preview: it shows the opening terms and the final term, and says whether the run grows, shrinks, stays constant or alternates in sign.
5. 05If you solved for the common ratio across an even number of steps, remember that the negative ratio of the same size fits your data equally well — the page shows the positive one.

The formula.

Getting from term 1 to term n takes n − 1 multiplications by r, so aₙ = a₁r^(n−1). The partial-sum formula comes from a one-line trick: write Sₙ, multiply the whole thing by r, subtract, and every term cancels except the first and the last, leaving Sₙ(1 − r) = a₁(1 − rⁿ) and therefore Sₙ = a₁(1 − rⁿ)/(1 − r). That derivation divides by 1 − r, so r = 1 has to be handled separately — with every term equal to a₁ the sum is just n × a₁, and this calculator branches on that case rather than letting it become 0/0. Letting n grow without bound, rⁿ tends to zero precisely when |r| < 1, which turns the partial-sum formula into S = a₁/(1 − r); outside that window rⁿ does not vanish and there is no limit. Rearranging the term formula gives the inverse solves: a₁ = aₙ/r^(n−1), and r = (aₙ/a₁)^(1/(n−1)). That root is where the ± ambiguity lives — with an even number of steps both signs work, and with an odd number only one does. Solving for the number of terms uses logarithms, n = 1 + ln|aₙ/a₁| ⁄ ln|r|, and because a logarithm never lands exactly on an integer in floating-point arithmetic, this page does not compare the log against an epsilon. It rounds the log to a candidate position, then recomputes a₁r^(candidate−1) and checks that against your aₙ at a relative tolerance of 1e-9 — which also enforces the sign parity when r is negative. Everything else is carried at full decimal precision and rounded once, at the end, to 10 decimal places; nothing is rounded at an intermediate step. Both verdicts on the page — whether the infinite series converges, and the growing/shrinking/constant/alternating word in the preview — are classified from the unrounded ratio, so a ratio of 0.999999999999 is still correctly called convergent even though the rounded output beside it reads 1.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), College Algebra 2e, §9.3 'Geometric Sequences' — 'Each term of a geometric sequence increases or decreases by a constant factor called the common ratio', and the explicit formula aₙ = a₁r^(n−1). Retrieved 2026-07-29; open access.
2. [2]OpenStax (Rice University), College Algebra 2e, §9.4 'Series and Their Notations' — the finite geometric series Sₙ = a₁(1 − rⁿ)/(1 − r) for r ≠ 1, and the infinite sum S = a₁/(1 − r) under the stated condition −1 < r < 1. Retrieved 2026-07-29; open access.
3. [3]Encyclopedia of Mathematics (European Mathematical Society), 'Geometric progression' — independent check: 'each one of which is equal to the preceding one multiplied by a number q ≠ 0', Sₙ = a₀(1 − qⁿ)/(1 − q), and 'If |q| < 1, the sum Sₙ tends to the limit S = a₀/(1 − q)'. Agrees with OpenStax on the worked example. Retrieved 2026-07-29; open access.
4. [4]Wolfram MathWorld, 'Geometric Series' — the closed form and the convergence condition, used as a second cross-check on the infinite-sum branch. Retrieved 2026-07-29; open access.
5. [5]OpenStax (Rice University), College Algebra 2e, §9.2 'Arithmetic Sequences' — cited for the additive/multiplicative contrast drawn throughout this page. Retrieved 2026-07-29; open access.