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Geometric sequences — the rules with examples

What is a geometric sequence and how do you find its terms?

In a geometric sequence the ratio between any two consecutive terms is constant — it is called the common ratio, r. Instead of adding the same number each time (as in a linear sequence), you multiply by the same number each time. From that constant ratio you can build a short formula for any term of the sequence, and even a formula for the sum of the terms — sometimes for the sum to infinity. At GCSE you need to recognise and continue a geometric sequence; the sum formulae are A level.

The three central formulae
nth term: aₙ = a₁ · rⁿ⁻¹ · sum: Sₙ = a₁(rⁿ − 1) ÷ (r − 1) · sum to infinity (when |r| < 1): S = a₁ ÷ (1 − r)
The geometric sequence 3, 6, 12, 24 — every bar is twice as tall as the one before, because the common ratio r = 2361224×2×2×2
In a geometric sequence every bar is r times the one before — here r = 2, so each bar doubles: 3, 6, 12, 24.

Worked examples, step by step

Recognising a geometric sequence and finding the nth term

The sequence is 3, 6, 12, 24. Show that it is geometric and find the nth term

  1. Check the ratio between consecutive terms: 6 ÷ 3 = 2, 12 ÷ 6 = 2, and 24 ÷ 12 = 2
  2. Every ratio equals 2, so this is a geometric sequence with r = 2
  3. Substitute into the nth-term formula: aₙ = a₁ · rⁿ⁻¹ = 3 · 2ⁿ⁻¹
  4. Check: the fourth term from the formula is a₄ = 3 · 2³ = 3 × 8 = 24 — it matches the given sequence
Finding a term from the formula

In a geometric sequence a₁ = 5 and r = 2. What is the sixth term?

  1. Substitute into the nth-term formula: aₙ = a₁ · rⁿ⁻¹
  2. For n = 6: a₆ = 5 · 2⁵
  3. Work out the power: 2⁵ = 32
  4. Answer: a₆ = 5 × 32 = 160
Finding r from two non-adjacent terms

In a geometric sequence a₂ = 6 and a₅ = 48. Find r and a₁

  1. Between the second and the fifth term there are 3 "jumps" of multiplying by r, so a₅ ÷ a₂ = r³
  2. Substitute: 48 ÷ 6 = 8 = r³, so r = 2
  3. Go back one step from a₂: a₁ = a₂ ÷ r = 6 ÷ 2 = 3
  4. Check: the sequence 3, 6, 12, 24, 48 — and the fifth term is indeed 48
The sum to infinity

Work out the sum 8 + 4 + 2 + 1 + ⋯

  1. Identify r: 4 ÷ 8 = 0.5, and check the pattern continues: 2 ÷ 4 = 0.5
  2. Because r = 0.5 and its size is less than 1, the sum to infinity exists
  3. Substitute into the sum-to-infinity formula: S = a₁ ÷ (1 − r) = 8 ÷ (1 − 0.5)
  4. Answer: S = 8 ÷ 0.5 = 16

Now try it yourself

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Formulae sheet
The nth term, the sum, and the differences between linear and geometric sequences.
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Learn sequences
From the first recognition to GCSE questions — explanation and guided practice.
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How to find the nth term
The other side of the coin — linear sequences step by step.

Frequently asked questions

What is the difference between a geometric sequence and a linear (arithmetic) sequence?

In a linear sequence you add the same constant difference at every step; in a geometric sequence you multiply by the same constant ratio r. 5, 8, 11 is linear (add 3); 5, 10, 20 is geometric (multiply by 2).

What happens when r is negative?

The signs of the terms alternate. For example with a₁ = 2 and r equal to −3, the terms are 2, −6, 18, −54 — each term comes from the previous one by multiplying by r, sign included.

When does a geometric sequence have a sum to infinity?

Only when the size of r is less than 1, that is r lies strictly between −1 and 1. Then each term is smaller than the one before, the terms "shrink" towards zero and the sum converges to a finite number S = a₁ ÷ (1 − r). If |r| is 1 or more, there is no finite sum.

How do you prove that a sequence is geometric?

Work out the ratio aₙ₊₁ ÷ aₙ from the nth term and show that the result is a constant that does not depend on n. That is a stronger argument than checking a few early terms, which can mislead.

✍️ Written by the MathsUK teamChecked against the National Curriculum and GCSE specificationsLast updated:

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