Practise 28 exam-style questions on Special sequences and geometric progressions for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
28 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
Here are the first four terms of an arithmetic sequence. 4.6, 7.9, 11.2, 14.5
Work out the 40th term of the sequence.
Model answer
Common difference \(= 3.3\) 40th term \(= 4.6 + 39 \times 3.3 = 4.6 + 128.7 = 133.3\)
Explanation
An arithmetic sequence adds the same amount each time, here 3.3. To reach the 40th term from the 1st term you add the difference 39 times, not 40 times. Alternatively, the \(n\)th term is \(3.3n + 1.3\).
Mark scheme
1 mark
M1 for \(4.6 + 39 \times 3.3\) or for the \(n\)th term \(3.3n + 1.3\)
oe; correct answer with no working scores full marks
1 mark
A1 for 133.3
cao
Question 2
2 marks
The sum of the first \(n\) cube numbers is equal to the square of the \(n\)th triangular number. For example, \(1^3 + 2^3 + 3^3 = 36 = 6^2\), and 6 is the 3rd triangular number.
Use this fact to work out \(1^3 + 2^3 + 3^3 + \ldots + 20^3\)
The \(n\)th triangular number is \(1 + 2 + \ldots + n = \frac{n(n + 1)}{2}\), so the 20th is 210. Squaring it gives the sum of the first 20 cubes without adding 20 separate cubes.
Mark scheme
1 mark
M1 for the 20th triangular number, 210
e.g. \(\dfrac{20 \times 21}{2}\) or \(1 + 2 + \ldots + 20\); correct answer with no working scores full marks
1 mark
A1 for 44 100
cao
Question 3
3 marks
Here are the first four terms of a geometric sequence. 128, 96, 72, 54
Which is the first term of the sequence that is less than 5? You must show your working.
Model answer
Common ratio \(= 96 \div 128 = 0.75\) \(n\)th term \(= 128 \times 0.75^{n - 1}\) 12th term \(= 128 \times 0.75^{11} = 5.406...\) (not less than 5) 13th term \(= 128 \times 0.75^{12} = 4.054...\) (less than 5) The 13th term is the first term less than 5.
Explanation
Find the common ratio by dividing a term by the previous one. The \(n\)th term of a geometric sequence is the first term × ratio\(^{n - 1}\), so you can test positions directly with powers rather than listing every term. Show the terms either side of 5 to justify your answer.
Mark scheme
1 mark
M1 for the common ratio 0.75
oe, e.g. \(\frac{3}{4}\)
1 mark
M1 for the 12th term (5.40...) and the 13th term (4.05...), e.g. \(128 \times 0.75^{11}\) and \(128 \times 0.75^{12}\)
cao; accept 4.05... with its position identified as the 13th; answer only scores 0
Question 4
5 marks
An arithmetic sequence and a geometric sequence both have first term 8 The common ratio of the geometric sequence is 1.25 The 3rd term of the arithmetic sequence is equal to the 3rd term of the geometric sequence.
Work out the difference between the 12th term of the geometric sequence and the 12th term of the arithmetic sequence. Give your answer correct to 1 decimal place.
Use the geometric sequence to find the shared 3rd term, then use it to find the arithmetic sequence's common difference (two steps from the 1st term to the 3rd term). The 12th terms are \(8 + 11d\) and \(8 \times 1.25^{11}\): there are 11 steps from the 1st term to the 12th. Growth by multiplying soon overtakes growth by adding.
Mark scheme
1 mark
P1 for the 3rd term of the geometric sequence, \(8 \times 1.25^2 = 12.5\)
correct answer with no working scores full marks
1 mark
P1 for the common difference of the arithmetic sequence, \(8 + 2d = 12.5\), so \(d = 2.25\)
exact
1 mark
P1 for the 12th term of the arithmetic sequence, \(8 + 11 \times 2.25 = 32.75\)
ft their \(d\)
1 mark
P1 for the 12th term of the geometric sequence, \(8 \times 1.25^{11} = 93.13...\)
accept 93.1
1 mark
A1 for 60.4
awrt 60.4 (60.382...)
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