Sequences & Series
22 free practice questions with explanations
PassNova has 22 free A-level Maths practice questions on Sequences & Series, each with a clear explanation. Practise them in the browser with instant feedback — 100% free, no sign-up, on any device. Updated for 2026.
Sequences & Series: example questions & answers
22 worked examples with answers and explanations below. Practise them in the browser with instant feedback on every answer.
An arithmetic series has first term a = 5 and common difference d = 3. Find the 20th term.
- A58
- B65
- C60
- D62✓
Answer: The nth term is a + (n − 1)d = 5 + 19 × 3 = 5 + 57 = 62.
A geometric sequence has first term 4 and common ratio 3. Find the 5th term.
- A324✓
- B81
- C108
- D972
Answer: The nth term is ar^(n−1) = 4 × 3⁴ = 4 × 81 = 324.
Find the sum to infinity of the geometric series 16 + 8 + 4 + 2 + …
- A24
- B30
- C32✓
- D64
Answer: With a = 16 and r = 1/2, the sum to infinity is a/(1 − r) = 16/(1 − 0.5) = 16/0.5 = 32.
Find the sum of the first 10 terms of the arithmetic series with a = 2 and d = 4.
- A180
- B220
- C110
- D200✓
Answer: Sₙ = (n/2)[2a + (n − 1)d] = 5 × [4 + 9 × 4] = 5 × 40 = 200.
The first three terms of an arithmetic sequence are 7, 11, 15. Find the 50th term.
- A199
- B211
- C200
- D203✓
Answer: With a = 7 and d = 4, the nth term is 7 + (n − 1)4, so the 50th term is 7 + 49 × 4 = 7 + 196 = 203.
An arithmetic sequence begins 3, 8, 13, 18, … Which term of the sequence is equal to 98?
- Athe 20th term✓
- Bthe 19th term
- Cthe 21st term
- Dthe 24th term
Answer: Here a = 3 and d = 5, so the nth term is 3 + (n − 1)5. Setting 3 + (n − 1)5 = 98 gives (n − 1)5 = 95, so n − 1 = 19 and n = 20.
Find the sum of the first 6 terms of the geometric series 2 + 6 + 18 + 54 + …
- A486
- B1458
- C728✓
- D364
Answer: With a = 2 and r = 3, Sₙ = a(rⁿ − 1)/(r − 1) = 2(3⁶ − 1)/(3 − 1) = 2(729 − 1)/2 = 728.
Find the sum to infinity of the geometric series 24 − 12 + 6 − 3 + …
- A48
- B16✓
- C12
- D36
Answer: Here a = 24 and r = −1/2 (since |r| < 1 the series converges). S∞ = a/(1 − r) = 24/(1 − (−1/2)) = 24/(3/2) = 16.
A sequence is defined by u₁ = 2 and uₙ₊₁ = 3uₙ − 1. Find the value of u₄.
- A13
- B14
- C122
- D41✓
Answer: Applying the recurrence: u₂ = 3(2) − 1 = 5, u₃ = 3(5) − 1 = 14, u₄ = 3(14) − 1 = 41.
Evaluate the sum Σ (from r = 1 to 5) of (2r + 1).
- A30
- B25
- C35✓
- D40
Answer: The terms for r = 1, 2, 3, 4, 5 are 3, 5, 7, 9, 11. Their sum is 3 + 5 + 7 + 9 + 11 = 35.
The geometric series 1 + 2x + (2x)² + (2x)³ + … converges to a finite sum. Find the range of values of x.
- A−1/2 < x < 1/2✓
- B−2 < x < 2
- C−1 < x < 1
- Dx > 1/2
Answer: A geometric series converges when |r| < 1. Here r = 2x, so |2x| < 1, giving |x| < 1/2, i.e. −1/2 < x < 1/2.
An arithmetic series has first term 4 and common difference 6. Find the sum of the first 30 terms.
- A2640
- B2730✓
- C5460
- D1365
Answer: Sₙ = (n/2)[2a + (n − 1)d] = (30/2)[2(4) + 29(6)] = 15[8 + 174] = 15 × 182 = 2730.
An arithmetic sequence has first term 5 and common difference 3. Find the sum of the first 20 terms.
- A670✓
- B62
- C620
- D1340
Answer: Sₙ = n/2[2a + (n − 1)d] = 20/2[2(5) + 19(3)] = 670. Giving the nth TERM rather than the sum is the classic misread of the question.
An arithmetic sequence has first term 100 and common difference −7. Find the sum of the first 15 terms.
- A765✓
- B2
- C15
- D1530
Answer: Sₙ = n/2[2a + (n − 1)d] = 15/2[2(100) + 14(−7)] = 765. Giving the nth TERM rather than the sum is the classic misread of the question.
An arithmetic sequence has first term 2 and common difference 4. Find the sum of the first 30 terms.
- A1770
- B118
- C1800✓
- D3600
Answer: Sₙ = n/2[2a + (n − 1)d] = 30/2[2(2) + 29(4)] = 1800. Giving the nth TERM rather than the sum is the classic misread of the question.
A geometric series has first term 3 and common ratio 1/2. Find the sum to infinity.
- A3/2
- B2
- C6✓
- D9/2
Answer: S∞ = a/(1 − r), valid because |r| < 1. Here S∞ = 3/(1 − 1/2) = 6. Using 1 + r instead of 1 − r is the usual slip.
A geometric series has first term 16 and common ratio 1/4. Find the sum to infinity.
- A64/5
- B64/3✓
- C4
- D20
Answer: S∞ = a/(1 − r), valid because |r| < 1. Here S∞ = 16/(1 − 1/4) = 64/3. Using 1 + r instead of 1 − r is the usual slip.
A geometric series has first term 5 and common ratio 2/3. Find the sum to infinity.
- A3
- B15✓
- C10/3
- D25/3
Answer: S∞ = a/(1 − r), valid because |r| < 1. Here S∞ = 5/(1 − 2/3) = 15. Using 1 + r instead of 1 − r is the usual slip.
An arithmetic sequence has first term 4 and common difference 6. Find the 10th term.
- A310
- B64
- C40
- D58✓
Answer: uₙ = a + (n − 1)d, so u₁₀ = 4 + 9(6) = 58. Using 10d instead of 9d is the off-by-one that catches most people — the first term already counts as one of the ten.
An arithmetic sequence has first term 11 and common difference −3. Find the 10th term.
- A−16✓
- B−19
- C110
- D−25
Answer: uₙ = a + (n − 1)d, so u₁₀ = 11 + 9(−3) = −16. Using 10d instead of 9d is the off-by-one that catches most people — the first term already counts as one of the ten.
A sequence is defined by u₁ = 3 and uₙ₊₁ = 2uₙ − 1. Find u₄.
- A23
- B15
- C17✓
- D9
Answer: Apply the rule repeatedly: u₂ = 2(3) − 1 = 5, u₃ = 2(5) − 1 = 9, u₄ = 2(9) − 1 = 17. A recurrence has to be iterated term by term — there is no shortcut to the fourth term here.
When does a geometric series have a sum to infinity?
- AWhen the number of terms in the series is sufficiently large
- BWhen the common ratio r is any positive number at all
- CWhen the first term of the series happens to be greater than zero
- DWhen |r| < 1✓
Answer: Only when successive terms shrink, which needs |r| < 1. If |r| ≥ 1 the terms do not tend to zero and the sum grows without limit, so S∞ = a/(1 − r) does not apply.