Geometric Sequence Formulas

You need a formula fast and cannot afford to dig through paragraphs of explanation. Here is every geometric sequence formula you need, with a short example and the exact conditions for each one. Sources: OpenStax Algebra and Trigonometry 2e, sections 13.3 and 13.4.

Formula	When To Use	Example
nth Term: aₙ = a₁ × rⁿ⁻¹	Find any term directly.	For 3, 6, 12, ... (a₁ = 3, r = 2), the 5th term is 3 × 2⁴ = 48.
Recursive Form: aₙ = r × aₙ₋₁	Define a term from the previous one.	For 3, 6, 12, ... a₅ = 2 × 12 = 24.
Sum of First n Terms: Sₙ = a₁ × (1 − rⁿ) / (1 − r), r ≠ 1	Add the first n terms.	For 3, 6, 12, 24, the sum of the first 4 terms is 3 × (1 − 2⁴) / (1 − 2) = 45.
Sum of First n Terms for r = 1: Sₙ = n × a₁	Every term is the same.	For 5, 5, 5, ... the sum of 4 terms is 4 × 5 = 20.
Sum to Infinity: S∞ = a₁ / (1 − r), |r| < 1	Add all terms forever when the series converges.	For 1, 0.5, 0.25, ... the infinite sum is 1 / (1 − 0.5) = 2.
Common Ratio From Two Terms: r = (aₙ / aₘ)^(1/(n−m))	Find r when you know two terms that are not consecutive.	If a₃ = 12 and a₅ = 48, then r = (48 / 12)^(1/2) = 2.
Find n With Logarithms: solve aₙ = a₁ × rⁿ⁻¹ for n	Find the position of a given term.	If a₁ = 3, r = 2, and aₙ = 48, then 48 = 3 × 2ⁿ⁻¹ → 16 = 2ⁿ⁻¹ → n − 1 = log₂(16) = 4 → n = 5.

This is the explicit formula for a geometric sequence. Use it to find any term directly without listing all the preceding ones.

Example: For the sequence 2, 6, 18, 54, ... the first term a₁ = 2 and the common ratio r = 3. The 7th term is a₇ = 2 × 3⁶ = 2 × 729 = 1458.

Edge case: The formula works for any real r except where r = 0, because the definition of a geometric sequence requires a non-zero common ratio (OpenStax 13.3). If r = 0, the sequence after the first term is all zeros, but it is excluded from the standard definition.

The recursive formula defines each term using the previous one: aₙ = r × aₙ₋₁. You also need the first term a₁.

Example: For a₁ = 2 and r = 3, the recursive form gives a₂ = 3 × 2 = 6, a₃ = 3 × 6 = 18, and so on.

Use the recursive formula when you are building a sequence step by step, or when a problem gives you a recurrence relation and asks for the explicit form. The recursive formula geometric sequence is also used as a starting point for proofs of the sum formulas.

Add a finite number of terms from a geometric sequence using the finite sum formula.

When r ≠ 1

Sₙ = a₁ × (1 − rⁿ) / (1 − r). This works for any r except 1.

Example: Calculate S₄ for 3, 6, 12, 24. a₁ = 3, r = 2, n = 4. S₄ = 3 × (1 − 16) / (1 − 2) = 3 × (−15) / (−1) = 45.

When r = 1

If r = 1, the formula above is undefined (division by zero). The correct formula is Sₙ = n × a₁. Every term is equal to a₁.

Example: For the constant sequence 5, 5, 5, 5, the sum of the first 4 terms is 4 × 5 = 20.

An infinite geometric series can have a finite sum, but only if the common ratio satisfies |r| < 1. This is the convergence condition for the infinite geometric series.

Formula: S∞ = a₁ / (1 − r).

Example: The series 1, 0.5, 0.25, 0.125, ... has a₁ = 1 and r = 0.5. The infinite sum is 1 / (1 − 0.5) = 2.

When |r| ≥ 1, the infinite sum diverges. For r = 1, the series is constant and grows without bound. For r = −1, the terms alternate (1, −1, 1, −1, ...) and the partial sums oscillate between 1 and 0, the series does not converge. For |r| > 1, terms grow in magnitude, so the sum is infinite.

Check the convergence condition before using the formula. A common mistake is to apply S∞ = a₁ / (1 − r) when r = −2, which gives a finite number for a divergent series.

When you know two terms of a geometric sequence but not the common ratio, use the common ratio formula: r = (aₙ / aₘ)^(1/(n−m)).

Example: a₃ = 12 and a₅ = 48. Here m = 3, n = 5, so r = (48 / 12)^(1/2) = 4^(1/2) = 2.

Failure case: If the two terms have opposite signs and n − m is even, the root can be ambiguous. For example, a₂ = −6 and a₄ = 24. The ratio of terms is 24 / (−6) = −4. Then r = (−4)^(1/2) = √(−4), which has no real solution. In this case, the sequence may not be geometric with a real ratio, or the data is inconsistent. Always check that the sign of aₙ / aₘ matches the expected sign of r^(n−m).

Sometimes you know a₁, r, and the value of a term aₙ, and need to find which position n it occupies. Solve aₙ = a₁ × rⁿ⁻¹ for n using logarithms.

Steps: Divide both sides by a₁: aₙ / a₁ = rⁿ⁻¹. Take the logarithm of both sides: log(aₙ / a₁) = (n − 1) × log(r). Then n = log(aₙ / a₁) / log(r) + 1.

Example: For a₁ = 2, r = 3, and aₙ = 162, find n. 162 / 2 = 81 = 3ⁿ⁻¹. Since 3⁴ = 81, n − 1 = 4, so n = 5.

Failure case: When solving for n from the partial sum formula Sₙ = a₁(1 − rⁿ)/(1 − r), the argument of the logarithm can become negative if a₁ and Sₙ(1 − r) have opposite signs. This yields no real solution. Check the sign before taking the logarithm.

The sum of geometric series formula for the first n terms is Sₙ = a₁(1 − rⁿ)/(1 − r). This is the same as the finite sum formula, but it is often written in sigma notation: Σ_{k=1}^{n} a₁ r^{k−1}.

Example: Σ_{k=1}^{4} 3 × 2^{k−1} = 3 + 6 + 12 + 24 = 45.

If the index starts at 0 instead of 1, the exponent changes. For Σ_{k=0}^{n−1} a₁ r^{k}, the formula is still Sₙ = a₁(1 − rⁿ)/(1 − r) because the first term is a₁ and the number of terms is n.

Copy this block for a one-page reference:

nth term: aₙ = a₁ × rⁿ⁻¹
Recursive: aₙ = r × aₙ₋₁
Finite sum (r ≠ 1): Sₙ = a₁ × (1 − rⁿ) / (1 − r)
Finite sum (r = 1): Sₙ = n × a₁
Infinite sum (|r| < 1): S∞ = a₁ / (1 − r)
Common ratio: r = (aₙ / aₘ)^(1/(n−m))
Find n: n = log(aₙ / a₁) / log(r) + 1

Check the convergence condition |r| < 1 before using the infinite sum formula. If r = 0, the sequence is not geometric by definition.