IB DP Mathematics: Analysis and Approaches – Sequences, series and financial applications Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the sequences, series and financial applications unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 1.1, 1.2, 1.3, 1.4 and 1.8, which are common content for SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 requires it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give answers exactly or to 3 significant figures unless a question says otherwise, and money to the nearest penny.

Learn the methods first in the study guide and the revision notes. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Give each answer in the form a × 10ᵏ, where 1 ≤ a < 10 and k ∈ ℤ.

(a) (3.6 × 10⁵) × (5 × 10⁻⁹) [2] (b) (8 × 10⁶) ÷ (2.5 × 10⁻²) [2]

2. (calculator-free) In an arithmetic sequence, u₄ = 17 and u₁₁ = 45. Find the common difference, the first term and the sum of the first 20 terms. [5]

3. (calculator-free) Evaluate the sum from r = 1 to 30 of (3r − 2). [3]

4. (calculator-free) An arithmetic sequence has first term 4 and common difference 6.

(a) Show that S_n = 3n² + n. [2] (b) Given that S_n = 990, find n. [3]

5. (calculator-free) A geometric sequence has u₂ = 48 and u₅ = −6.

(a) Find the common ratio. [2] (b) Find u₁. [1] (c) Explain why the sum to infinity exists, and find it. [3]

6. (calculator-free) Consider the geometric series 1 + (2x − 1) + (2x − 1)² + …

(a) Find the set of values of x for which the series converges. [2] (b) Given that the sum to infinity is 4, find x. [3]

7. (calculator allowed) A sports club records its membership at the start of five consecutive years.

Year	1	2	3	4	5
Members	212	238	261	289	312

(a) Show that the membership does not form an exact arithmetic sequence. [1] (b) Using the first and last values, find an approximate common difference. [2] (c) Use your answer to (b) to predict the membership in year 8. [1]

8. (calculator allowed) Amira invests £12 000 in an account paying 4.2% annual interest, compounded monthly.

(a) Find the value of the investment after 8 years. [2] (b) Inflation is 2.5% per year throughout. Find the real value of the investment after 8 years, in today’s money. [2] (c) Find the number of complete years after which the value first exceeds £20 000. [3]

9. (calculator allowed) A car is bought for £28 500. Its value depreciates by 15% each year.

(a) Find its value after 5 years. [2] (b) Find the number of complete years after which its value first falls below £10 000. [2]

10. (calculator allowed) In a model of an outbreak, 40 new cases are reported on day 1, and each day there are 18% more new cases than the day before.

(a) Find the number of new cases on day 14. [2] (b) Write the total number of cases over the first 14 days in sigma notation, and find this total to the nearest whole number. [2] (c) Find the first day on which the number of new cases exceeds 1000. [2]

11. (calculator allowed) A ball is dropped from a height of 5 m. After each bounce it rises to 60% of the height from which it last fell. Assume it keeps bouncing in this way indefinitely.

(a) Find the height the ball reaches after its 4th bounce. [2] (b) Find the first bounce after which the ball reaches a height of less than 0.1 m. [3] (c) Find the total vertical distance the ball travels. [3]

12. (calculator allowed) Two job offers pay an annual salary. Offer A pays £42 000 in year 1 and rises by £1800 each year. Offer B pays £40 000 in year 1 and rises by 5% each year.

(a) Find the salary in year 10 for each offer. [3] (b) Find the total earned over the first 10 years for each offer. [4] (c) Find the first year in which the salary from offer B is greater than the salary from offer A. [2] (d) Find the least number of complete years after which the total earned from offer B is greater than the total earned from offer A. [2]

Answers

1. (a) (3.6 × 5) × 10⁵⁻⁹ = 18 × 10⁻⁴ [1] = 1.8 × 10⁻³ [1] (b) (8 ÷ 2.5) × 10⁶⁻⁽⁻²⁾ [1] = 3.2 × 10⁸ [1] Examiner insight: 18 × 10⁻⁴ is correct in value but not in the required form, so it earns the method mark only; 3.2E8 is calculator notation and is not accepted.

2. u₁₁ − u₄ = 7d = 28 [1], so d = 4 [1]. u₁ = 17 − 3 × 4 = 5 [1]. S₂₀ = (20/2)(2 × 5 + 19 × 4) [1] = 10 × 86 = 860 [1] Examiner insight: The method mark for S₂₀ is awarded for correct substitution into the sum formula, so it can still be earned with a wrong d carried through on follow-through.

3. 30 terms, first term 1, last term 3(30) − 2 = 88 [1]. Sum = (30/2)(1 + 88) [1] = 1335 [1] Examiner insight: Stating the number of terms, first term and last term earns the first mark even if the arithmetic that follows slips.

4. (a) S_n = (n/2)(2 × 4 + (n − 1) × 6) [1] = (n/2)(6n + 2) = n(3n + 1) = 3n² + n [1] (b) 3n² + n − 990 = 0 [1]; (n − 18)(3n + 55) = 0 [1]; n = −55/3 is rejected as n must be a positive integer, so n = 18 [1] Examiner insight: In a “show that”, the line (n/2)(6n + 2) must appear; jumping from the formula to 3n² + n loses the second mark.

5. (a) u₅/u₂ = r³ = −6/48 = −1/8 [1], so r = −1/2 [1] (b) u₁ = 48 ÷ (−1/2) = −96 [1] (c) |r| = 1/2 < 1, so the series converges [1]. S∞ = −96/(1 − (−1/2)) [1] = −96/(3/2) = −64 [1] Examiner insight: “Explain why” needs the statement |r| < 1 with the value of r; writing “because r is small” does not earn the reasoning mark.

6. (a) |2x − 1| < 1 [1], so −1 < 2x − 1 < 1, giving 0 < x < 1 [1] (b) 1/(1 − (2x − 1)) = 4 [1]; 1/(2 − 2x) = 4, so 2 − 2x = 1/4 [1]; x = 7/8, which lies in 0 < x < 1 [1] Examiner insight: Non-strict inequalities (0 ≤ x ≤ 1) lose the accuracy mark in (a), because the series diverges at both end points.

7. (a) The differences are 26, 23, 28, 23, which are not constant, so the sequence is not exactly arithmetic [1] (b) d ≈ (312 − 212)/4 [1] = 25 members per year [1] (c) u₈ ≈ 212 + 7 × 25 = 387 members [1] Examiner insight: In (a) you must show the differences; “the numbers do not go up evenly” with no values is not enough for the mark.

8. (a) FV = 12 000 × (1 + 0.042/12)⁹⁶ [1] = £16 782.22 [1] (b) Real value = 16 782.22… ÷ 1.025⁸ [1] = £13 773.95 [1] (c) 12 000 × 1.0035¹²ⁿ > 20 000 [1]. After 12 years the value is £19 846.48; after 13 years it is £20 696.27 [1]. So 13 years [1] Examiner insight: If you use TVM, write N = 96, I% = 4.2, PV = −12 000, P/Y = C/Y = 12; a bare final answer that is wrong earns nothing, but recorded inputs can earn the method mark.

9. (a) 28 500 × 0.85⁵ [1] = £12 645.60 [1] (b) After 6 years £10 748.76; after 7 years £9136.45 [1], so 7 years [1] Examiner insight: A multiplier of 0.15 instead of 0.85 is a method error, so no follow-through accuracy mark is available in (a).

10. (a) u₁₄ = 40 × 1.18¹³ [1] = 343.97…, so 344 new cases [1] (b) Total = Σ from n = 1 to 14 of 40(1.18)ⁿ⁻¹ [1] = 40(1.18¹⁴ − 1)/0.18 = 2032.72…, so 2033 cases [1] (c) 40 × 1.18ⁿ⁻¹ > 1000 gives n − 1 > ln 25/ln 1.18 = 19.45 [1], so n > 20.45… and the first whole day is day 21 [1] Examiner insight: Using 1.18¹⁴ for day 14 is the most common index slip; day n uses the power n − 1 because day 1 has no increase applied.

11. (a) 5 × 0.6⁴ [1] = 0.648 m [1] (b) 5 × 0.6ⁿ < 0.1 [1]; n > ln 0.02/ln 0.6 = 7.66 [1]; so the 8th bounce [1] (c) Distance = 5 + 2(3 + 1.8 + 1.08 + …) [1] = 5 + 2 × 3/(1 − 0.6) [1] = 5 + 15 = 20 m [1] Examiner insight: The first drop of 5 m is travelled once, but every later height is travelled twice (up and down); missing the factor 2 or counting 5 m twice loses the final accuracy mark.

12. (a) A: 42 000 + 9 × 1800 = £58 200 [1]. B: 40 000 × 1.05⁹ [1] = £62 053.13 [1] (b) A: (10/2)(2 × 42 000 + 9 × 1800) [1] = £501 000 [1]. B: 40 000(1.05¹⁰ − 1)/0.05 [1] = £503 115.70 [1] (c) Using a GDC table: year 5 gives £48 620.25 < £49 200, year 6 gives £51 051.26 > £51 000 [1]. Year 6 [1] (d) After 9 years the totals are £441 062.57 (B) and £442 800 (A); after 10 years £503 115.70 (B) and £501 000 (A) [1]. 10 years [1] Examiner insight: In (c) and (d) the method mark needs the values either side of the change; a bare wrong answer earns nothing.

Where marks are usually lost

• Leaving an answer as 18 × 10⁻⁴ or 1.8E−3 rather than 1.8 × 10⁻³.
• Miscounting terms in a sigma sum, especially when the lower limit is not 1.
• Using rⁿ instead of rⁿ⁻¹ when u₁ is the day 1 or year 1 value.
• Using the annual rate per period: 4.2% compounded monthly means 0.35% per month over 12n months.
• Rounding a future value before dividing by the inflation factor, which moves the real value by a penny or more.
• Rounding n down in “first exceeds” questions, or not showing the values either side of the boundary.
• Writing only a GDC answer with no inputs or table values, so no method mark can be given.
• Forgetting to check |r| < 1 before using S∞, or giving non-strict inequalities for convergence.
• Keeping the negative root when solving a quadratic in n.
• Counting the first drop twice (or later bounces once) in total-distance problems.

Next steps

• Recap the formulas: revision notes
• Relearn a method: study guide
• Course overview: IB DP Maths AA course hub
• Track your coverage: printable syllabus checklist
• Plan your final weeks: IB DP Maths AA exam preparation guide
• Check other subjects with all free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020).