Math · Pre-Calculus

Geometric sequences

Q: Find the 6th term of the geometric sequence 3, 6, 12, 24, ...

96. Common ratio r=2, first term a₁=3: a₆ = 3 × 2⁵ = 3×32 = 96. 48 (B) comes from using the exponent (n−2) instead of (n−1) — computing 3×2⁴ = 48, one power short.

Q: In a geometric sequence, the 2nd term is 6 and the 5th term is 162. Find the common ratio.

3. Divide the 5th term by the 2nd term to eliminate a₁: a₅/a₂ = r³ = 162/6 = 27, so r = 3. 27 (B) is r³ itself — stopping before taking the cube root to isolate r.

Q: Find the sum of the infinite geometric sequence 8, −4, 2, −1, ...

16/3. The common ratio is r = −4/8 = −1/2, and since |r| < 1, the infinite sum converges: S = a₁/(1−r) = 8/(1−(−1/2)) = 8/(3/2) = 16/3. 16 (B) comes from using r = +1/2 instead of −1/2 — missing that the sequence alternates sign, which changes the denominator from 3/2 to 1/2 and inflates the result.

Find a term, the common ratio, or the sum of a sequence that changes by a constant multiple each step — the same structure as an arithmetic sequence, but multiplication replaces addition throughout. Every question comes with a written walkthrough of which formula applied.

Practise geometric sequences nowSee sample questions

Difficulty tiers

35s

Target pace

55%

Average first-attempt score

What's covered

The nth term formula — apply aₙ = a₁ · r^(n−1), where a₁ is the first term and r is the common ratio between consecutive terms.
Finding the common ratio — divide any term by the one right before it — the result should be the same no matter which consecutive pair is used.
The sum of a finite geometric sequence — apply Sₙ = a₁(1−rⁿ)/(1−r) for r≠1, rather than adding every term individually.
The sum of an infinite geometric sequence — apply S = a₁/(1−r), but only when |r|<1 — an infinite sum with |r|≥1 doesn't converge to a finite value at all.
Recognizing a geometric sequence — check that the ratio between every pair of consecutive terms is constant, not just the difference.

Where students lose marks

Confusing the exponent's off-by-one with arithmetic sequences

Just like the arithmetic nth term formula, the exponent on r is (n−1), not n — plugging in n directly shifts every term by one position.

Applying the infinite sum formula when |r|≥1

If the common ratio's absolute value is 1 or greater, the terms don't shrink toward zero, and the sum grows without bound — there's no finite value to compute.

Mixing up r and 1/r

Dividing a term by the one before it, rather than the one after it, gives 1/r instead of r — every downstream calculation using the flipped ratio comes out wrong.

Three sample questions

Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.

Sample 01Foundation

Find the 6th term of the geometric sequence 3, 6, 12, 24, ...

Sample 02Core

In a geometric sequence, the 2nd term is 6 and the 5th term is 162. Find the common ratio.

Sample 03Advanced

Find the sum of the infinite geometric sequence 8, −4, 2, −1, ...

How to practise this

Geometric sequences are arithmetic sequences with multiplication in place of addition — the same formula structure, the same off-by-one risks, plus a convergence condition to check on infinite sums.

Check the ratio and the convergence condition before summing

Confirm the ratio is constant, and for an infinite sum, check |r|<1 before applying the formula at all. Untimed practice is where that check becomes automatic.

Then 35 seconds a question

Move to timed sessions once ratio-finding and the nth-term formula are instant. Infinite sums are quick once the convergence check is a reflex.

Fold into a mock

Pair geometric sequences with arithmetic sequences in a mock — recognizing which formula a sequence needs is the real skill being tested.

Stop guessing the ratio.
Start checking it first.

A formula walkthrough on every item
Timed or untimed sessions, any length you like
Tracked separately — see your geometric sequences accuracy and pace over time

Start practising free

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