How to find the nth term of a geometric sequence

To find the nth term of a geometric sequence, use aₙ = a₁ × r^(n-1). That is the only thing you need for most homework problems, but knowing when and how to apply it requires you to find the common ratio r and the first term a₁ first. The standard homework question types are solved step by step: given two terms, find n; given a term value, find its position; and how to handle growth and decay word problems.

When you are given two non-consecutive terms of a geometric sequence, the common ratio r is the key to everything else. Suppose you know aₘ and aₙ, with m < n. The ratio of the two terms equals r raised to the difference in their positions: aₙ / aₘ = r^(n-m). To isolate r, take the (n-m)th root: r = (aₙ / aₘ)^(1/(n-m)).

Watch the Sign When Taking an Even Root

If n-m is even and the ratio aₙ / aₘ is negative, you get two possible real roots, one positive, one negative. Check the sequence: if terms alternate sign, r is negative; if they stay the same sign, r is positive. For example, given a₂ = 4 and a₆ = -64, the ratio is -64/4 = -16, and n-m = 4. The fourth root of -16 is not a real number, so the given terms cannot belong to the same geometric sequence with real r. That is the correct answer: no real sequence exists.

Find the First Term After Finding r

Once you have r, plug any known term into the explicit expression to find a₁. Use aₙ = a₁ × r^(n-1), so a₁ = aₙ / r^(n-1). If r is negative and n-1 is odd, the sign carries through correctly.

OpenStax Algebra and Trigonometry 2e, section 13.3, defines a geometric sequence as one in which the ratio of any term to the previous term is constant, and it gives the explicit expression aₙ = a₁ × r^(n-1) as the standard tool for finding any term.

When you know the term value aₙ and need to find its position n, set up aₙ = a₁ × r^(n-1) and solve for n. Divide both sides by a₁: aₙ / a₁ = r^(n-1). Take the logarithm base r of both sides: log_r(aₙ / a₁) = n-1. Then n = log_r(aₙ / a₁) + 1. If your calculator does not have a log base r button, use the change-of-base trick: log_r(x) = ln(x) / ln(r).

What Goes Wrong

The argument of the logarithm must be positive. If aₙ / a₁ is negative and r is negative, the equality still holds for some n, but the logarithm of a negative number is undefined in real numbers. In that case, you must check sign patterns manually: if r is negative and aₙ and a₁ have opposite signs, n-1 must be odd; if they share the same sign, n-1 must be even. Solve for the integer n directly by testing powers.

For example, in the sequence 6, -12, 24, -48, ..., find the position of the term 384. Here a₁ = 6 and r = -2. Set 384 = 6 × (-2)^(n-1). Divide: 64 = (-2)^(n-1). Since 64 is positive and r is negative, n-1 must be even. (-2)^6 = 64, so n-1 = 6 and n = 7. The 7th term is 384.

To sum the first n terms of a geometric sequence, use the partial sum equation Sₙ = a₁ × (1 - rⁿ) / (1 - r) when r ≠ 1. Work step by step: compute rⁿ, subtract from 1, divide by (1 - r), then multiply by a₁. If r = 1, every term equals a₁, so Sₙ = n × a₁. OpenStax section 13.4 gives both equations.

The Edge Case That Traps Students

When a₁ and (Sₙ × (1 - r)) have opposite signs, the argument 1 - rⁿ inside the equation can become negative, and solving for n using logarithms gives no real solution. This happens when the partial sum is smaller than the first term for a decaying sequence. For instance, in the sequence 100, 50, 25, ..., the sum after 2 terms is 150, but after 3 terms it is 175. If a problem asks for n such that Sₙ = 80, no real n exists because the sum never drops below 100. The correct answer is that there is no solution.

In growth problems, the common ratio r is greater than 1: each term multiplies the previous one by a factor that represents a percentage increase. In decay problems, r is between 0 and 1: each term is a percentage of the previous one. The same expressions apply; the only difference is the context of the numbers.

Growth Example

A bacteria culture starts with 500 cells and triples every hour. How many cells after 6 hours? Here a₁ = 500, r = 3, n = 7 (because the first hour is hour 0, so after 6 hours you want term 7). a₇ = 500 × 3⁶ = 500 × 729 = 364,500 cells.

Decay Example

A radioactive substance has a half-life of 10 years. If you start with 200 grams, how much remains after 40 years? a₁ = 200, r = 0.5, and after 40 years you are at n = 5 (since 40/10 = 4 half-lives, so term 5). a₅ = 200 × 0.5⁴ = 200 × 0.0625 = 12.5 grams.

For both growth and decay, check whether the problem asks for the term value or the sum. If the question is about total accumulation (like total bacteria after 6 hours), use the partial sum equation. If it asks for the amount at a specific time, use the nth term expression.

Students make the same five errors repeatedly. Memorise this checklist:

• Using the arithmetic expression: Plugging into aₙ = a₁ + (n-1)d instead of aₙ = a₁ × r^(n-1). The geometric sequence multiplies; the arithmetic sequence adds.
• Sign error when finding r from two terms: Taking the wrong root when n-m is even and the term ratio is negative. Always check whether the terms alternate sign.
• Logarithm of a negative number: Forgetting that the argument of a logarithm must be positive. When the sequence alternates sign, solve by testing powers manually.
• Using the sum equation for r = 1: The equation Sₙ = a₁ × (1 - rⁿ) / (1 - r) is undefined at r = 1. Use Sₙ = n × a₁ instead.
• Applying the infinite sum equation when |r| ≥ 1: The infinite geometric series only converges when |r| < 1. If |r| ≥ 1, the series diverges and has no finite sum. Stewart Calculus, section 11.2, states this convergence condition explicitly.

The most common single error is confusing r with |r| for convergence: students write that a geometric series converges when r < 1, but r < -1 makes |r| > 1 and the series diverges. Always check the absolute value.

This material suits Algebra 2 and precalculus students who need to find terms, the common ratio, or the number of terms from given information. It also suits calculus students who need to determine whether an infinite geometric series converges and, if so, compute its sum. Teachers can use the worked examples as an answer key or demonstration of the relationship between the sequence expression and the series sum equation.

Anyone who needs to compute the sum of an arithmetic series should use an arithmetic series calculator instead. Students looking for proofs of convergence tests other than the ratio test for geometric series should consult a calculus textbook section on the ratio test.