OCR GCSE Mathematics: Sequences, nth term and quadratic sequences (J560) – 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 cover sections 6.06a and 6.06b (sequences) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), first assessed in 2017. Questions 8 to 11 carry the label (Higher tier only), since the specification lists their content in the Higher tier column. Question 12 uses Foundation and Higher content only. Learn the methods first in the study guide or the revision notes. For planning, use the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. The first term of a sequence is 3. The rule is “multiply by 4, then subtract 6”. Write down the next three terms. [2]

2. The nth term of a sequence is 6n − 5.

(a) Write down the first three terms. [1] (b) Work out the 40th term. [1] (c) Show that 100 is not a term of this sequence. [2]

3. Here is a sequence: 23, 19, 15, 11, …

(a) Work out a formula for the nth term. [2] (b) Which is the first term below −50? Give its value. [3]

4. A gardener lays paving stones in a path. Path 1 uses 5 stones, path 2 uses 8 stones and path 3 uses 11 stones. Each path is made by adding a column of 3 stones to the one before.

(a) Write a formula for the number of stones in path n. [2] (b) One path uses 128 stones. Which path is it? [2]

5. (a) Write down the 8th triangular number. [1] (b) From 45, 64, 100, 125, 216, write down all the cube numbers. [2] (c) Show that the sum of the 3rd and 4th triangular numbers is a square number. [1]

6. A Fibonacci-type sequence has 3rd term 13 and 6th term 57. Find the first two terms and write down the first six terms. [4]

7. (a) Find the common ratio of the geometric progression 625, 375, 225, … and write down the next two terms. [3] (b) Decide whether 3, 6, 11, 18, … is arithmetic, geometric or quadratic. Give a reason and write down the next term. [3]

8. (Higher tier only) A sequence has x₁ = 5 and xₙ₊₁ = 2xₙ − 7.

(a) Find x₂, x₃ and x₄. [3] (b) A different sequence has the same rule xₙ₊₁ = 2xₙ − 7 and x₃ = 11. Find its first term x₁. [3]

9. (Higher tier only) Here is a sequence: 6, 13, 24, 39, 58, …

(a) Work out the nth term. [4] (b) Work out the 20th term. [1] (c) Show that 500 is not a term of the sequence. [2]

10. (Higher tier only) Find the nth term of 3, 7, 12, 18, 25, … Give your answer in the form an² + bn. [4]

11. (Higher tier only)

(a) Find the nth term of 1/4, 2/7, 3/10, 4/13, … [2] (b) Find the nth term of √5, √10, √15, √20, … Write the 4th term in the form a√5, and find the 20th term. [3] (c) The geometric sequence 3, 3√2, 6, 6√2, … has common ratio √2. Write down its nth term and find the 6th term in the form a√2. [3]

12. Sequence A has nth term 5n + 2. Sequence B has nth term 40 − 3n.

(a) Write down the first three terms of each sequence. [2] (b) Show that no term of A is equal to the term of B in the same position. [2] (c) Find the first position in which the term of A is greater than the term of B. [2]

Answers

1. 2nd term: 4 × 3 − 6 = 6 [1]. 3rd term: 4 × 6 − 6 = 18, 4th term: 4 × 18 − 6 = 66 [1]. The terms are 6, 18, 66. Examiner insight: Show each term on its own line. A slip early on carries through the rest, and written working lets you see where it happened.

2. (a) n = 1, 2, 3 gives 6 − 5, 12 − 5, 18 − 5, so 1, 7, 13. [1] (b) 6 × 40 − 5 = 240 − 5 = 235. [1] (c) Set 6n − 5 = 100, so 6n = 105 and n = 17.5 [1]. The position must be a whole number, so 100 is not a term. [1] Examiner insight: A “show that” answer needs the equation solved and a sentence linking the non-whole n to the conclusion. A bare “no” does not show anything.

3. (a) d = −4 [1]. The terms of −4n are −4, −8, … and 23 is 27 more than −4, so the nth term is 27 − 4n. [1] (b) Solve 27 − 4n < −50, so 4n > 77 and n > 19.25 [1]. The first whole n is 20 [1]. The 20th term is 27 − 80 = −53. [1] Examiner insight: Give the term, not only the position. Check your answer: the 19th term is −49, which is not below −50, so the 20th is the first.

4. (a) The common difference is 3, so start with 3n [1]. 3n gives 3, 6, 9 and each path has 2 more, so 3n + 2. [1] (b) 3n + 2 = 128 gives 3n = 126 [1], so n = 42, which is path 42. [1] Examiner insight: In context questions, use the structure of the diagram to explain your formula. Here 3n is the columns of 3 and the 2 is the part that does not change.

5. (a) 8 × 9/2 = 36. [1] (b) 64 = 4³ [1] and 125 = 5³ and 216 = 6³ [1]. The cube numbers are 64, 125 and 216. (c) 3rd triangular number is 6 and 4th is 10, and 6 + 10 = 16 = 4², a square number. [1] Examiner insight: Write down the numbers you are adding, not only “16”. The line 6 + 10 = 16 is what shows the claim holds.

6. Call the first two terms a and b. The 3rd term gives a + b = 13 [1]. The 6th term is 3a + 5b = 57 [1]. Substituting a = 13 − b: 39 − 3b + 5b = 57, so 2b = 18, b = 9 [1]. Then a = 4. The sequence is 4, 9, 13, 22, 35, 57. [1] Examiner insight: Check by adding forwards: 13 + 22 = 35 and 22 + 35 = 57. If the check fails, the algebra has gone wrong.

7. (a) r = 375 ÷ 625 = 3/5 [1]. Check: 225 ÷ 375 = 3/5. Next terms: 225 × 3/5 = 135 [1] and 135 × 3/5 = 81. [1] (b) Differences 3, 5, 7. Not constant, so not arithmetic [1]. Ratios 6 ÷ 3 = 2 and 11 ÷ 6 = 11/6 differ, so not geometric. Second differences are 2, 2, so it is quadratic [1]. Next first difference 9, next term 18 + 9 = 27. [1] Examiner insight: Give a reason for the type: name the differences or ratios you tested. A label with no evidence leaves nothing to judge.

8. (a) x₂ = 2 × 5 − 7 = 3 [1]. x₃ = 2 × 3 − 7 = −1 [1]. x₄ = 2 × (−1) − 7 = −9. [1] (b) x₃ = 2x₂ − 7 = 11, so 2x₂ = 18 and x₂ = 9 [1]. x₂ = 2x₁ − 7 = 9, so 2x₁ = 16 [1]. x₁ = 8. [1] Examiner insight: Be careful with the negative in x₄. Write 2 × (−1) in brackets so that −2 − 7 gives −9.

9. (a) First differences 7, 11, 15, 19 and second differences 4, 4, 4 [1]. So a = 4 ÷ 2 = 2 [1]. Terms of 2n²: 2, 8, 18, 32, 50. Subtracting from the sequence gives 4, 5, 6, 7, 8 [1], which is n + 3. The nth term is 2n² + n + 3. [1] (b) 2 × 400 + 20 + 3 = 823. [1] (c) 2n² + n + 3 = 500 gives 2n² + n − 497 = 0 [1]. The discriminant is 1 + 8 × 497 = 3977, which lies between 63² = 3969 and 64² = 4096, so it is not a perfect square and n is not a whole number. [1] Examiner insight: Check your formula on the first two terms before using it in (b) and (c). A wrong formula spoils all three parts.

10. First differences 4, 5, 6, 7 and second differences 1, 1, 1 [1]. So a = 1/2 [1]. Terms of n²/2: 0.5, 2, 4.5, 8, 12.5. Subtracting: 2.5, 5, 7.5, 10, 12.5 [1], which is 2.5n. The nth term is n²/2 + 5n/2, which equals n(n + 5)/2. [1] Examiner insight: When the second difference is odd, a is a fraction. Keep it as a fraction rather than rounding, and the remainder will then be linear.

11. (a) Numerators 1, 2, 3, 4 give n [1]. Denominators 4, 7, 10, 13 increase by 3, so 3n + 1. The nth term is n/(3n + 1). [1] (b) The numbers under the root are 5, 10, 15, 20, so the nth term is √(5n) [1]. The 4th term is √20 = 2√5 [1]. The 20th term is √100 = 10. [1] (c) The first term is 3 and each term multiplies by √2, so the nth term is 3 × (√2)ⁿ⁻¹ [1]. For n = 6: 3 × (√2)⁵ [1]. (√2)⁵ = 4√2, so the 6th term is 12√2. [1] Examiner insight: For a fraction sequence, treat the numerator and the denominator as two separate sequences. For surds, simplify the final answer fully.

12. (a) A: 7, 12, 17. B: 37, 34, 31. [2] (b) Set 5n + 2 = 40 − 3n, so 8n = 38 [1]. n = 4.75, which is not a whole number, so there is no position where they are equal. [1] (c) A is greater when 5n + 2 > 40 − 3n, so n > 4.75 [1]. The first whole position is 5. Check: A = 27 and B = 25. [1] Examiner insight: An inequality gives a range of n, so round to a whole position and check it. Testing positions 4 and 5 directly confirms the answer.

Where marks are usually lost

• Using the common difference as the constant, so 6n − 5 becomes n − 5.
• Dropping the sign on a decreasing sequence, writing 27 + 4n for 27 − 4n.
• Giving a decimal n as a position instead of concluding that the number is not a term.
• Mixing up arithmetic and geometric: adding the ratio or multiplying by the difference.
• Calling a sequence quadratic without computing the second differences.
• Not halving the second difference before finding the n² coefficient.
• Skipping the check on the first two terms of a quadratic nth term.
• Leaving a surd answer unsimplified, such as √20 instead of 2√5.
• Treating xₙ₊₁ as xₙ + 1.
• Answering with the position when the question asks for the term.

Next steps

• Review the methods in the revision notes.
• Read the full explanations in the study guide.
• Browse other topics on the course hub and record progress on the checklist.
• A free 10-minute diagnostic will show your weakest topics.
• Book a free trial class if you want a teacher to go through your answers.

Official syllabus

Source: OCR, GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), section 6.06 Sequences (6.06a and 6.06b).