S5 Maths sequences paper 1: nth term and partial sums
- 2026-11-10
S5 Maths sequences paper 1: nth term and partial sums
Ten items start from the basic arithmetic and geometric formulas, then build partial sums and reverse questions.
Direct answers Items 01 to 03 use the nth term formula, 04 to 06 partial sums and solving for n, 07 to 09 reverse problems on the difference or ratio, and 10 the mistake of treating a geometric sequence as arithmetic.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Maths Arithmetic and Geometric Sequences |
| Level | S5 |
02 Questions (10 items)
- FoundationAn arithmetic sequence has first term 3 and common difference 4. Find the 10th term.
- FoundationThe sequence 5, 8, 11, … is arithmetic. Find the 12th term.
- FoundationA geometric sequence has first term 2 and common ratio 3. Find the 5th term.
- IntermediateAn arithmetic sequence has first term 7 and common difference 3. Find the sum of the first 15 terms.
- IntermediateFor the geometric sequence 3, 6, 12, … find the 8th term.
- IntermediateAn arithmetic sequence has first term 12 and common difference -3. Which term first equals 0?
- ChallengeAn arithmetic sequence has first term 7 and its first three terms add to 33. Find the 10th term.
- ChallengeA geometric sequence has first term 5 and common ratio 2. Find the sum of the first 6 terms.
- ChallengeIn an arithmetic sequence the 3rd term is 11 and the 7th term is 27. Find the 15th term.
- Error redoError drill: a student treated 2, 6, 18, … as arithmetic and got 26 for the 5th term. What is the correct 5th term?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 39 | The nth term is a + (n – 1)d, which gives 3 + 9 times 4, so 39. |
| 2 | 38 | The common difference is 3, so the 12th term is 5 + 11 times 3, which is 38. |
| 3 | 162 | The nth term is a times r to the power n – 1, so 2 times 3 to the fourth power, which is 162. |
| 4 | 420 | The sum is 15/2 times (14 + 42), which is 7.5 times 56, so 420. |
| 5 | 384 | The common ratio is 2, so the 8th term is 3 times 2 to the seventh power, which is 384. |
| 6 | The 5th term | Solving 12 + (n – 1)(-3) = 0 gives 3n = 15, so n = 5. |
| 7 | 43 | The three-term sum gives 21 + 3d = 33, so d = 4 and the 10th term is 7 + 36 = 43. |
| 8 | 315 | The sum is 5 times (2 to the sixth power minus 1) divided by 1, which is 5 times 63, so 315. |
| 9 | 59 | The four steps add 16, so the common difference is 4 and the first term is 3; the 15th term is 3 + 56 = 59. |
| 10 | 162 | The sequence has common ratio 3, so the 5th term is 2 times 3 to the fourth power, which is 162; 26 comes from wrongly adding 6 each time. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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