OCR A Level Mathematics A: Sequences, sigma notation, arithmetic and geometric series and modelling (H240) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018), sections 1.04e to 1.04k. All of this content is A Level only (Stage 2). Learn the method from the study guide or revision notes first. The course page is OCR A Level Mathematics and the printable list is the checklist.

Questions

1. A sequence has uₙ = n² − 8n + 20. (a) Write down u₁ to u₅. [1] (b) Explain why the sequence is neither increasing nor decreasing. [1] (c) Find the values of n for which uₙ = 13. [2]

2. A sequence is defined by xₙ₊₁ = 0.6xₙ + 8 with x₁ = 5. Find x₂ and x₃, then find the limit L that the terms approach, justifying your method. [3]

3. A sequence has x₁ = 5 and xₙ₊₁ = k − xₙ, where k is a constant. The sum of the first 30 terms is 135. Find k and state the order of the sequence. [4]

4. (a) Write 5 + 8 + 11 + … + 32 using sigma notation. [2] (b) Evaluate Σ (4r − 7), from r = 5 to r = 30. [3]

5. The sum of the first n terms of a sequence is Sₙ = 3n² + 2n. (a) Find u₁. [1] (b) Show that uₙ = 6n − 1 and that the sequence is arithmetic. [3] (c) Find u₁₀₀. [1]

6. A geometric sequence has second term 54 and fifth term 16. (a) Find the common ratio and the first term. [3] (b) Find the sum to infinity, stating why it exists. [2]

7. A geometric series has first term 10 and common ratio (3 − x)/5. (a) Find the values of x for which the series converges. [2] (b) Find the sum to infinity when x = −1. [2]

8. A saver pays £300 into an account at the start of each year. The account earns 4.5% compound interest per year. Find the total in the account at the end of the 12th year, to the nearest pound. [4]

9. A machine costs £24 000 and loses 12% of its value each year. (a) Find its value after 3 years. [1] (b) Find the first year in which its value is below £5000. [4]

10. Show that the recurring decimal 0.1353535… can be written as 67/495. [4]

11. A loan of £20 000 is charged 5% interest per year. At the end of each year, after interest is added, a payment of £x is made. Let Bₙ be the amount owed after n payments. (a) Show that Bₙ = 20 000 × 1.05ⁿ − x(1.05ⁿ − 1)/0.05. [3] (b) Find x, to the nearest penny, so that the loan is cleared by the 10th payment. [3] (c) Find the total repaid, to the nearest pound. [1]

12. Two training plans start in week 1. Plan A runs 5 km in week 1 and adds 0.8 km each week. Plan B runs 4 km in week 1 and increases the distance by 12% each week. (a) Find the distance in week 10 for each plan, to 2 decimal places. [2] (b) Show that week 12 is the first week in which Plan B’s distance is greater than Plan A’s. [3] (c) Find the total distance for each plan over the first 12 weeks, to 1 decimal place. [3]

Answers

1. (a) 13, 8, 5, 4, 5 [1] (b) The terms fall from u₁ to u₄ and then rise, so uₙ₊₁ − uₙ is not one sign for all n. [1] (c) n² − 8n + 20 = 13 gives n² − 8n + 7 = 0 [1], so (n − 1)(n − 7) = 0 and n = 1 or n = 7 [1] Examiner insight: A list of terms shows a change of direction; naming the term where the direction changes makes the explanation complete.

2. x₂ = 0.6(5) + 8 = 11 [1] and x₃ = 0.6(11) + 8 = 14.6 [1]. At the limit xₙ₊₁ and xₙ both equal L, so L = 0.6L + 8, 0.4L = 8 and L = 20 [1] Examiner insight: Writing the limit equation L = f(L) is the step that shows the method; quoting 20 from a long list of terms is weaker.

3. The terms are 5, k − 5, 5, k − 5, … so each pair sums to k [1]. There are 15 pairs in 30 terms [1], so 15k = 135 [1]. Then k = 9 and the order is 2 [1] Examiner insight: The key move is to group the terms; explaining why the sequence repeats avoids relying on a guess.

4. (a) The rth term is 3r + 2 [1]. Since 32 = 3(10) + 2, the series is Σ (3r + 2), r = 1 to 10 [1] (b) The number of terms is 30 − 5 + 1 = 26 [1], first term 13, last term 113 [1]. S = 26/2 × (13 + 113) = 1638 [1] Examiner insight: Checking both end terms of the sigma expression is a good habit, and counting terms as last minus first plus one avoids an off-by-one slip.

5. (a) u₁ = S₁ = 3 + 2 = 5 [1] (b) For n ≥ 2, uₙ = Sₙ − Sₙ₋₁ [1] = 3n² + 2n − 3(n − 1)² − 2(n − 1) = 6n − 1 [1]. This also gives 5 when n = 1, and uₙ₊₁ − uₙ = 6 is constant, so the sequence is arithmetic [1] (c) u₁₀₀ = 600 − 1 = 599 [1] Examiner insight: In a “show that” the final form is given, so every algebraic step in the expansion should be written; also check n = 1 separately.

6. (a) r³ = 16/54 = 8/27 [1], so r = 2/3 [1]. Then a × (2/3) = 54 gives a = 81 [1] (b) |r| = 2/3 < 1, so the series converges [1]. S∞ = 81/(1 − 2/3) = 243 [1] Examiner insight: Stating |r| < 1 before using the sum to infinity formula shows you have checked that the series converges.

7. (a) |3 − x|/5 < 1, so |3 − x| < 5 [1], giving −2 < x < 8 [1] (b) r = 4/5 [1] and S∞ = 10/(1 − 4/5) = 50 [1] Examiner insight: A modulus condition gives a double inequality; a single inequality such as x < 8 alone usually loses the lower bound.

8. The first payment earns 12 years of interest, the last earns 1 year [1]. The total is 300(1.045) + 300(1.045)² + … + 300(1.045)¹², a geometric series with a = 313.5 and r = 1.045 [1]. S₁₂ = 313.5(1.045¹² − 1)/0.045 [1], so the total is £4848 [1] Examiner insight: Listing the payments with their powers first shows the structure; jumping to a formula with the wrong first term is the usual slip.

9. (a) 24 000 × 0.88³ = £16 355.33 [1] (b) 24 000 × 0.88ⁿ < 5000 [1], so 0.88ⁿ < 5/24 and n ln 0.88 < ln(5/24) [1]. Dividing by ln 0.88, which is negative, reverses the inequality: n > 12.27… [1]. The first whole year is year 13 [1] Examiner insight: Dividing by a negative logarithm is a common place to lose a mark; checking the values at 12 and 13 years confirms the direction.

10. 0.1353535… = 0.1 + 0.035 + 0.00035 + 0.0000035 + … [1]. The terms after 0.1 form a geometric series with a = 0.035 and r = 0.01 [1]. Its sum is 0.035/0.99 = 7/198 [1]. Over the common denominator 990, 1/10 = 99/990 and 7/198 = 35/990, so the total is 134/990 = 67/495 [1] Examiner insight: A “show that” needs every step; combine the fractions over a common denominator and simplify visibly.

11. (a) B₁ = 1.05(20 000) − x [1]. B₂ = 1.05B₁ − x = 20 000 × 1.05² − x(1.05 + 1), and so on, so Bₙ = 20 000 × 1.05ⁿ − x(1 + 1.05 + … + 1.05ⁿ⁻¹) [1]. The bracket is a geometric series with sum (1.05ⁿ − 1)/0.05, giving the result [1] (b) B₁₀ = 0 [1], so x = 20 000 × 1.05¹⁰ × 0.05/(1.05¹⁰ − 1) [1] = £2590.09 [1] (c) 10 × 2590.09 = £25 901 [1] Examiner insight: Show how the previous balance is multiplied and reduced before generalising; stating the geometric sum separately makes the derivation easy to follow.

12. (a) Plan A: 5 + 9(0.8) = 12.20 km [1]. Plan B: 4 × 1.12⁹ = 11.09 km [1] (b) Plan A is 5 + 0.8(w − 1) and Plan B is 4 × 1.12^(w−1) [1]. Week 11: A = 13.00, B = 12.42 [1]. Week 12: A = 13.80, B = 13.91, so B is first greater [1] (c) Plan A: 12/2 × (2 × 5 + 11 × 0.8) = 112.8 km [1]. Plan B: 4(1.12¹² − 1)/0.12 [1] = 96.5 km [1] Examiner insight: Comparing both values at the two weeks either side of the crossing point is what proves “first”; one value alone does not.

Where marks are usually lost

• Counting the number of terms in a sigma sum as last minus first instead of last minus first plus one.
• Using the sum to infinity formula without checking |r| < 1.
• Leaving a modulus condition as a single inequality.
• Forgetting to reverse an inequality when dividing by a negative logarithm.
• Using 0.12 instead of 0.88 as the ratio for a 12% fall.
• Rounding an interest ratio early, so the final pound figure is out.
• Giving n as a decimal when the question asks for a whole number of years or weeks.
• Assuming the first term of a saving plan earns interest for the wrong number of periods.

Next steps

• Revise from the revision notes.
• Re-read the study guide.
• Course page: OCR A Level Mathematics and the checklist.
• Try the free diagnostics to find your weakest topics.
• Book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.04e to 1.04k.