OxfordAQA International GCSE Mathematics: Sequences and nth terms (9260) – 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 references A24, A25 and A26 in section 3.2.4 (Sequences) of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Questions 9 to 12 are Extension only because they use quadratic sequences. Questions 1 to 8 are Core content, which can also be assessed on the Extension tier question papers.

Read the study guide or the revision notes first if you need to. The course page is OxfordAQA IGCSE Mathematics, the printable list is the checklist, and you can also take a free 10-minute diagnostic.

Questions

1. A sequence starts at 3. To find each new term, double the previous term and then add 1.

(a) Write down the 2nd, 3rd and 4th terms. [2] (b) Work out the 5th term. [1]

2. The nth term of a sequence is 9n − 5.

(a) Write down the first three terms. [1] (b) Work out the 10th term. [1] (c) Is 301 a term of this sequence? You must show your working. [2]

3. Here is a list of numbers: 10, 16, 27, 28, 49, 64, 91, 125.

(a) Which numbers in the list are triangular? [2] (b) Which numbers in the list are cube numbers but not square numbers? [1]

4. Here are the first four terms of a sequence: 20, 17, 14, 11, …

(a) Find an expression for the nth term. [2] (b) Find the first term of the sequence that is negative. [2]

5. The 5th term of an arithmetic progression is 31. The 12th term is 66.

(a) Work out the common difference. [2] (b) Find an expression for the nth term. [2] (c) Show that 200 is not a term of the sequence. [1]

6. A runner follows a training plan. She runs 2.5 km in week 1. In each later week she runs 0.75 km more than in the week before.

(a) Show that the distance, in km, she runs in week n is 0.75n + 1.75. [2] (b) In which week does the distance first exceed 15 km? [3]

7. The nth triangular number is n(n + 1)/2.

(a) Work out the 15th triangular number. [1] (b) The nth triangular number is 253. Find n. [3]

8. The 7th and 8th triangular numbers are 28 and 36.

(a) Show that the sum of the 7th and 8th triangular numbers is a square number. [2] (b) Show that the sum of the nth and (n + 1)th triangular numbers is (n + 1)². [2]

9. (Extension only) The first five terms of a quadratic sequence are 7, 14, 27, 46 and 71. Find the nth term. [4]

10. (Extension only) The nth term of a sequence is 2n² − 5n + 1.

(a) Write down the first four terms. [2] (b) Show that 89 is a term of the sequence and state its position. [3]

11. (Extension only) Tiles are used to make a set of patterns. The numbers of tiles in the first five patterns are 4, 10, 18, 28, 40. The numbers follow a quadratic sequence.

(a) Find an expression for the number of tiles in pattern n. [3] (b) Which pattern uses exactly 130 tiles? [3]

12. (Extension only) A quadratic sequence has nth term an² + bn + c. The second difference of the sequence is 4. The 2nd term is 9 and the 3rd term is 21.

(a) Find the values of a, b and c. [4] (b) Find the first term of the sequence that is greater than 500. [3]

Answers

1. (a) 2 × 3 + 1 = 7 [1]. 2 × 7 + 1 = 15 and 2 × 15 + 1 = 31. The terms are 7, 15, 31 [1]. (b) 2 × 31 + 1 = 63 [1]. Examiner insight: Show each application of the rule on its own line. Writing every line makes a slip easy to spot and correct.

2. (a) 9(1) − 5 = 4, 9(2) − 5 = 13, 9(3) − 5 = 22. The terms are 4, 13, 22 [1]. (b) 9(10) − 5 = 85 [1]. (c) Solve 9n − 5 = 301, so 9n = 306 [1] and n = 34. This is a whole number, so yes, 301 is the 34th term [1]. Examiner insight: “Is it a term?” needs a conclusion as well as a value of n. Say that n is a whole number and name the position.

3. (a) The triangular numbers run 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91 and so on [1]. In the list these are 10, 28 and 91 [1]. (b) The cubes in the list are 27 = 3³, 64 = 4³ and 125 = 5³. Of these, 64 = 8² is also a square, so the answer is 27 and 125 [1]. Examiner insight: Write out the family up to the largest number in the list rather than relying on memory. Larger members, such as 91, are the easiest to miss, and you can test one with n(n + 1) = 182 = 13 × 14.

4. (a) The common difference is −3 [1]. The first term is 20, so the constant is 20 + 3 = 23 and the nth term is 23 − 3n [1]. (b) Solve 23 − 3n < 0, so n > 7.67 [1]. The first whole number is 8, and the 8th term is 23 − 24 = −1 [1]. Examiner insight: A reader can see that you have used the sign of the common difference correctly when you write −3n first. Check your formula by substituting n = 1.

5. (a) From the 5th term to the 12th term there are 7 steps and the terms rise by 66 − 31 = 35 [1]. So d = 35 ÷ 7 = 5 [1]. (b) The nth term is 5n + c. Using the 5th term, 25 + c = 31, so c = 6 [1]. The nth term is 5n + 6 [1]. Check: 5(12) + 6 = 66. (c) Solve 5n + 6 = 200, so n = 38.8 [1]. A position has to be a whole number, so 200 is not in the sequence. Examiner insight: On a “show that” question the final line must state the conclusion. Give the value of n, say it is not a whole number, and say what that means.

6. (a) The common difference is 0.75 [1]. The nth term is 0.75n + c, and week 1 gives 0.75 + c = 2.5, so c = 1.75. The nth term is 0.75n + 1.75 [1]. (b) Solve 0.75n + 1.75 > 15, so 0.75n > 13.25 [1] and n > 17.67 [1]. Week 17 gives 14.5 km, which is not enough. Week 18 gives 0.75(18) + 1.75 = 15.25. The answer is week 18 [1]. Examiner insight: Round the way the context requires. A week is a whole number, and checking the weeks either side of the boundary shows that your answer is the first week.

7. (a) 15 × 16 ÷ 2 = 120 [1]. (b) n(n + 1)/2 = 253, so n(n + 1) = 506 [1]. Rearranging gives n² + n − 506 = 0, which factorises as (n − 22)(n + 23) = 0 [1]. A position cannot be negative, so n = 22 [1]. Examiner insight: Rejecting n = −23 explicitly shows that you have thought about the context. Alternatively, spotting 22 × 23 = 506 is fine if you show the product.

8. (a) 28 + 36 = 64 [1], and 64 = 8², which is a square number [1]. (b) The sum is n(n + 1)/2 + (n + 1)(n + 2)/2 = (n + 1)(n + n + 2)/2 [1]. This equals (n + 1)(2n + 2)/2 = (n + 1)(n + 1) = (n + 1)² [1]. Examiner insight: A “show that” needs every step of algebra. Taking out the common factor (n + 1) before simplifying keeps the working short and clear.

9. First differences: 7, 13, 19, 25. Second differences: 6, 6, 6 [1]. So a = 3 [1]. The terms of 3n² are 3, 12, 27, 48, 75. Sequence − 3n²: 4, 2, 0, −2, −4, which is 6 − 2n [1]. The nth term is 3n² − 2n + 6 [1]. Check: n = 2 gives 12 − 4 + 6 = 14. Examiner insight: Set the differences out in a table, and test your formula on a term you did not use to build it, for example the 4th term, which is 46.

10. (a) n = 1: 2 − 5 + 1 = −2. n = 2: 8 − 10 + 1 = −1 [1]. n = 3: 18 − 15 + 1 = 4. n = 4: 32 − 20 + 1 = 13. The terms are −2, −1, 4, 13 [1]. (b) Solve 2n² − 5n + 1 = 89, so 2n² − 5n − 88 = 0 [1]. Factorising gives (2n + 11)(n − 8) = 0 [1]. The root n = −5.5 is rejected, so n = 8. 89 is the 8th term [1]. Check: 2(64) − 40 + 1 = 89. Examiner insight: Quote the position as well as showing that n is a whole number. A negative or fractional root should be rejected in words.

11. (a) First differences: 6, 8, 10, 12. Second differences: 2, 2, 2, so a = 1 [1]. Sequence − n²: 3, 6, 9, 12, 15, which is 3n [1]. The nth term is n² + 3n [1]. (b) Solve n² + 3n = 130, so n² + 3n − 130 = 0 [1]. Factorising gives (n − 10)(n + 13) = 0 [1]. Pattern number 10 uses 130 tiles, since n = −13 is rejected [1]. Examiner insight: Part (b) depends on your answer to part (a). Showing the equation you form from your own formula makes your reasoning clear to the reader.

12. (a) The second difference is 2a = 4, so a = 2 [1]. The 2nd term gives 2(4) + 2b + c = 9, so 2b + c = 1 [1]. The 3rd term gives 2(9) + 3b + c = 21, so 3b + c = 3 [1]. Subtracting gives b = 2, then c = −3. So a = 2, b = 2, c = −3 [1]. (b) The nth term is 2n² + 2n − 3. The 15th term is 450 + 30 − 3 = 477, which is not enough [1]. The 16th term is 512 + 32 − 3 = 541 [1]. The first term greater than 500 is the 16th term, 541 [1]. Examiner insight: Write the two simultaneous equations clearly before solving them. In part (b), trying the terms either side of 500 proves that you have found the first one.

Where marks are usually lost

• Writing the first term as the coefficient of n instead of the common difference.
• Losing the sign of a negative common difference, for example writing 3n + 23 instead of 23 − 3n.
• Not checking the formula against the first two terms of the sequence.
• Treating a decimal or negative value of n as a valid position.
• Rounding to the nearest whole number instead of the next whole number up in “first term greater than” questions.
• Using the second difference as a, instead of halving it.
• Subtracting the wrong terms when finding the linear remainder in a quadratic sequence.
• Stopping at “n = 22” without saying which term it is, or without answering the question that was asked.
• Missing steps in “show that” questions, especially the algebra with brackets.
• Confusing triangular numbers with square numbers, or forgetting that some numbers are both.

Next steps

• Review the revision notes and the study guide.
• Return to the OxfordAQA IGCSE Mathematics course page and tick off topics on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.4, references A24, A25 and A26.