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IB Maths AA: Sequences and Series (10 Basic Items)
IB Maths AA: Sequences and Series (10 Basic Items)
Ten basic items on arithmetic terms and sums, the nth term of a geometric sequence, the sum to infinity, counting terms, and the classic ar^n versus ar^(n−1) confusion.
Focus Tell arithmetic from geometric, and always start the index at n − 1.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 30 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: IB Maths AA: Sequences and Series |
| Level | IB Maths AA (SL) |
02 Questions (10 items)
- FoundationAn arithmetic sequence has first term 3 and common difference 5. Find the 10th term.
- FoundationA geometric sequence has first term 2 and common ratio 3. Find the 6th term.
- FoundationFind the sum to infinity of 8 + 4 + 2 + 1 + …
- FoundationHow many terms are there in the arithmetic sequence 5, 9, 13, …, 97?
- IntermediateAn arithmetic sequence has first term 4 and common difference 6. Find the sum of the first 12 terms.
- IntermediateFor the geometric sequence 5, 10, 20, …, find the 8th term and the sum of the first 8 terms.
- IntermediateA geometric series has first term 12 and common ratio 0.8. Find its sum to infinity and explain why the sum exists.
- ChallengeAn arithmetic sequence has first term 6 and common difference 4. Which term equals 50?
- ChallengeFor the same sequence (first term 3, common difference 5), find the sum of the first 20 terms.
- Error redoA student writes “the nth term is ar^n, so with a = 3 and r = 2 the first term is 6”. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | u10 = 3 + (10 − 1) × 5 = 3 + 45 = 48. | The nth term is a + (n − 1)d, so the difference is added 9 times, not 10. |
| 2 | u6 = 2 × 3^5 = 2 × 243 = 486. | The nth term of a geometric sequence is ar^(n − 1): the index is one less than the term number. |
| 3 | a = 8 and r = 1/2, so S = 8 ÷ (1 − 1/2) = 16. | The formula only holds when |r| < 1, which is satisfied here because r = 1/2. |
| 4 | (97 − 5) ÷ 4 + 1 = 24 terms. | Number of terms = (last − first) ÷ common difference + 1; the final plus one is easily forgotten. |
| 5 | S12 = 12/2 × [8 + 66] = 6 × 74 = 444. | Use S = n/2[2a + (n − 1)d], working out the bracket before multiplying by n/2. |
| 6 | u8 = 5 × 2^7 = 640; S8 = 5(2^8 − 1) ÷ (2 − 1) = 5 × 255 = 1275. | The numerator is (r^n − 1); writing it this way avoids negatives when r > 1. |
| 7 | S = 12 ÷ (1 − 0.8) = 60, and the series converges because |0.8| < 1. | Check |r| < 1 first, otherwise the formula has no meaning. |
| 8 | 6 + (n − 1) × 4 = 50, so (n − 1) × 4 = 44 and n = 12. | To find a term number, set the general term equal to the target and solve; do not count term by term. |
| 9 | S20 = 10 × [6 + 95] = 10 × 101 = 1010. | Even though the 10th term is known, the sum still needs a and d rather than one term value. |
| 10 | The first term is ar^0 = 3 × 1 = 3. Using ar^n multiplies every term by r and makes the whole sequence wrong. | The index starts at zero; mixing this up is the commonest error with the general term. |
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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