AQA GCSE Mathematics 8300: Sequences – 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 section 3.2.4 Sequences (A23, A24 and A25) of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Questions and parts on content from the specification’s “Higher content only” column are labelled (Higher); everything else is for both tiers. Each question is labelled (non-calculator), in the style of Paper 1, or (calculator), in the style of Papers 2 and 3.

Learn the content first in the Sequences study guide and the Sequences revision notes. These questions are different from the sequences items in the topic-level Algebra practice questions, so you can do both. The course hub is AQA GCSE Mathematics and the printable checklist lists every statement.

Questions

1. (non-calculator) The nth term of a sequence is 5n − 8.

(a) Write down the first three terms of the sequence. [1] (b) Work out the 40th term. [1]

2. (non-calculator) The first term of a sequence is 4. The term-to-term rule is “multiply by 3, then subtract 5”. Work out the 2nd, 3rd and 4th terms. [2]

3. (non-calculator) Matchsticks are used to make a row of hexagons. Next to each other, the hexagons share a side. Pattern 1 uses 6 matchsticks, pattern 2 uses 11 and pattern 3 uses 16.

(a) Work out how many matchsticks are in pattern 10. [1] (b) Find an expression for the number of matchsticks in pattern n. [2] (c) Can a pattern use exactly 100 matchsticks? Show how you decide. [2]

4. (non-calculator) Here is a list of numbers: 3, 8, 9, 21, 25, 27, 36, 64

(a) Write down all the triangular numbers in the list. [1] (b) Write down the number in the list that is both a square number and a cube number. [1] (c) Write down the next triangular number after 36. [1]

5. (non-calculator) In a Fibonacci-type sequence, each term is the sum of the two terms before it. The 3rd term is 13 and the 6th term is 59. Work out the first two terms. [4]

6. (calculator) Here are the first four terms of a geometric progression: 2500, 1000, 400, 160, …

(a) Work out the common ratio. [1] (b) Work out the next two terms. [1] (c) Which term is the first one less than 1? [2]

7. (non-calculator) Here are the first four terms of a linear sequence: 47, 41, 35, 29, …

(a) Find an expression for the nth term. [2] (b) Work out the first term of the sequence that is less than −40. [2]

8. (non-calculator) Sequence A has nth term 4n + 9. Sequence B has nth term 7n − 12. For one value of n, the nth terms of the two sequences are equal. Work out this value of n and the term. [3]

9. (non-calculator) Here are the first five terms of a quadratic sequence: 3, 4, 7, 12, 19, …

(a) Work out the next two terms. [2] (b) (Higher) Find an expression for the nth term. [3]

10. (non-calculator, Higher) A geometric progression begins 3, 3√2, 6, …

(a) Write down the common ratio. [1] (b) Work out the 8th term. Give your answer in the form k√2. [2] (c) Which term is equal to 96? [2]

11. (calculator) Each pattern is a square of white tiles with a border of grey tiles around it. Pattern 1 has 1 white tile and 8 grey tiles. Pattern 2 has 4 white tiles and 12 grey tiles. Pattern 3 has 9 white tiles and 16 grey tiles. The pattern continues.

(a) Find an expression for the number of grey tiles in pattern n. [2] (b) Write down the number of white tiles in pattern 10. [1] (c) Show that the total number of tiles in pattern n is (n + 2)². [2] (d) One pattern has 140 grey tiles. Work out the number of white tiles in this pattern. [2] (e) (Higher) Show that no pattern has the same number of white tiles as grey tiles. [2]

12. (non-calculator) A sequence has first term 2. Each term after that is found using the rule

next term = 1 ÷ (1 − previous term)

(a) Work out the 2nd, 3rd and 4th terms. [2] (b) (Higher) Work out the 50th term. Explain how you know. [2] (c) (Higher) Work out the sum of the first 50 terms. [3]

Answers

1. (a) 5 − 8, 10 − 8, 15 − 8 give −3, 2, 7 [1] (b) 5 × 40 − 8 = 192 [1] Examiner insight: The mark in (a) needs all three terms correct, so a sign slip on the negative first term loses it; write each substitution so the slip is easy to spot.

2. 3 × 4 − 5 = 7 and 3 × 7 − 5 = 16 [1]; 3 × 16 − 5 = 43, so 7, 16, 43 [1] Examiner insight: An error in the 2nd term is usually followed through for the later terms if the rule is applied correctly, so keep going rather than starting again.

3. (a) The sequence goes up by 5: 6, 11, 16, 21, … Pattern 10 = 6 + 9 × 5 = 51 [1] (b) Common difference 5, so start with 5n: 5, 10, 15 [1]; each term is 1 more, so 5n + 1 [1] (c) 5n + 1 = 100 gives 5n = 99, n = 19.8 [1]; n is not a whole number, so no [1] Examiner insight: In (b), an answer of “n + 5” scores nothing, while “5n + k” with the wrong k often earns the first mark; in (c) the final mark needs the reason, not just “no”.

4. (a) 3, 21, 36 [1] (b) 64 (8² = 64 and 4³ = 64) [1] (c) 36 + 9 = 45 [1] Examiner insight: For a “write down all” part, including a wrong number (such as 9 or 25 in part (a)) usually loses the mark even if every correct number is there.

5. Let the first two terms be a and b. Then the terms are a, b, a + b, a + 2b, 2a + 3b, 3a + 5b [1] a + b = 13 and 3a + 5b = 59 [1] 3a + 3b = 39, so subtracting gives 2b = 20, b = 10 [1] a = 13 − 10 = 3, so the first two terms are 3 and 10 [1] (check: 3, 10, 13, 23, 36, 59) Examiner insight: Trial and improvement can reach 3 and 10, but if it fails it earns nothing; correct expressions for the later terms earn method marks even when the solving goes wrong.

6. (a) 1000 ÷ 2500 = 0.4 [1] (b) 160 × 0.4 = 64, 64 × 0.4 = 25.6 [1] (c) Continuing: 10.24, 4.096, 1.6384 (9th), 0.65536 (10th) [1]; the first term less than 1 is the 10th term [1] Examiner insight: The method mark in (c) needs the terms listed with their positions; the answer is a position, so giving “0.65536” alone answers a different question.

7. (a) Common difference −6, so −6n: −6, −12, −18 [1]; each term is 53 more, so 53 − 6n [1] (b) 53 − 6n < −40 gives 6n > 93, n > 15.5 [1]; the 16th term is 53 − 96 = −43 [1] Examiner insight: “6n + 53” or “53 + 6n” gets only the mark for the constant, since the sign of n is the common difference; in (b) stopping at n = 16 does not answer “work out the term”.

8. 4n + 9 = 7n − 12 [1] 21 = 3n, so n = 7 [1] 4 × 7 + 9 = 37 (and 7 × 7 − 12 = 37) [1] Examiner insight: Listing terms of both sequences until they meet is acceptable if every term is correct, but the equation method is quicker and gets the method mark straight away.

9. (a) First differences 1, 3, 5, 7; second difference 2 [1]; next first differences 9 and 11, so 28 and 39 [1] (b) Second difference 2, so a = 1: start with n² [1] Term − n²: 3 − 1, 4 − 4, 7 − 9, 12 − 16, … = 2, 0, −2, −4, which is −2n + 4 [1] nth term n² − 2n + 4 [1] (check n = 5: 25 − 10 + 4 = 19) Examiner insight: The first mark is for the correct n² coefficient, which is half the second difference; writing 2n² usually means no further marks can be earned on the linear part.

10. (a) 3√2 ÷ 3 = √2 [1] (b) Terms: 3, 3√2, 6, 6√2, 12, 12√2, 24, 24√2, so 3 × (√2)⁷ [1] = 3 × 8√2 = 24√2 [1] (c) Every second term doubles: 6 (3rd), 12 (5th), 24 (7th), 48 (9th), 96 [1]; 96 is the 11th term [1] Examiner insight: A decimal such as 33.9 for (b) does not meet “the form k√2”, so the accuracy mark is lost even with correct method.

11. (a) Common difference 4, so 4n: 4, 8, 12 [1]; each term is 4 more, so 4n + 4 [1] (b) White tiles are square numbers, n², so 10² = 100 [1] (c) Total = n² + 4n + 4 [1] = (n + 2)(n + 2) = (n + 2)² [1] (d) 4n + 4 = 140, so n = 34 [1]; white tiles = 34² = 1156 [1] (e) n² = 4n + 4 gives n² − 4n − 4 = 0 [1] n = (4 ± √32)/2 = 2 ± 2√2, about 4.83 or −0.83; neither is a positive whole number, so no such pattern [1] Examiner insight: In (c) a “show that” needs the expansion of (n + 2)² or the sum n² + 4n + 4 shown in full; checking one pattern number only does not prove it for all n.

12. (a) 1 ÷ (1 − 2) = −1 and 1 ÷ (1 − (−1)) = 1/2 [1]; 1 ÷ (1 − 1/2) = 2, so −1, 1/2, 2 [1] (b) The terms repeat in a cycle of 3: 2, −1, 1/2, 2, −1, 1/2, … [1]; 50 = 3 × 16 + 2, so the 50th term is the 2nd in the cycle: −1 [1] (c) Each cycle sums to 2 − 1 + 1/2 = 3/2 [1] 16 cycles give 16 × 3/2 = 24, then terms 49 and 50 are 2 and −1 [1] Sum = 24 + 2 − 1 = 25 [1] Examiner insight: A bare “−1” in (b) does not earn the explanation mark; you must show the repeating cycle and link 50 to its position in the cycle.

Where marks are usually lost

• Writing the term-to-term rule (“add 5”) instead of the nth term when the question asks for an expression in n.
• Using the first term as the coefficient of n, as in “6n + 5” for 6, 11, 16, …
• Leaving out the reason when deciding whether a number is in a sequence: show the equation and the value of n.
• Answering a “which term” question with the term’s value, or a “work out the term” question with the position.
• Losing the sign on a decreasing sequence.
• Including extra numbers in a “write down all” list question.
• Using the full second difference as the n² coefficient instead of half of it (Higher).
• Giving surd answers as decimals when an exact form is asked for (Higher).
• Not showing the repeating pattern when a recursive sequence cycles (Higher).

Next steps

• Sequences revision notes
• Sequences study guide
• Algebra study guide for the rest of Topic 2
• AQA GCSE Mathematics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Mathematics (8300) specification, version 1.0, for teaching from September 2015 onwards and exams from May/June 2017 onwards, published by AQA. Section 3.2.4 Sequences, references A23 to A25.