Special sequences and geometric progressions - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 29 exam-style questions on Special sequences and geometric progressions for Edexcel GCSE Maths (Non-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.

29 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.
7, 11, 15, 19

Work out the 50th term of the sequence.

Model answer

The common difference is 4.
The 50th term is 49 steps on from the first term.
\(7 + 49 \times 4 = 7 + 196 = 203\)

Explanation

In an arithmetic sequence the same amount is added each time. To reach the 50th term from the 1st you add the difference 49 times, not 50. Equivalently, the \(n\)th term is \(4n + 3\).

Mark scheme

1 mark
M1 for a correct method, e.g. \(7 + 49 \times 4\) or the \(n\)th term \(4n + 3\)
correct answer with no working scores full marks
1 mark
A1 for 203
cao

Question 2

2 marks

Here are the first five terms of a sequence.
1, 3, 7, 15, 31

Write down an expression, in terms of \(n\), for the \(n\)th term of the sequence.

Model answer

Powers of 2: 2, 4, 8, 16, 32
Each term is 1 less than these.
\(n\)th term \(= 2^n - 1\)

Explanation

The differences 2, 4, 8, 16 double each time, which suggests powers of 2. Comparing the terms with \(2^n\) shows each one is exactly 1 less. Check: \(2^5 - 1 = 31\).

Mark scheme

1 mark
M1 for linking the terms to powers of 2, e.g. 2, 4, 8, 16, 32 or each term is 1 less than a power of 2
correct answer with no working scores full marks
1 mark
A1 for \(2^n - 1\)
oe

Question 3

3 marks

A geometric sequence has first term 1 and common ratio \(\sqrt{5}\)

Which term of the sequence is equal to 625?

Model answer

The \(n\)th term is \((\sqrt{5})^{n-1}\).
\(625 = 5^4 = (\sqrt{5})^8\)
\(n - 1 = 8\), so \(n = 9\)
(Terms: 1, \(\sqrt{5}\), 5, \(5\sqrt{5}\), 25, \(25\sqrt{5}\), 125, \(125\sqrt{5}\), 625)
It is the 9th term.

Explanation

Since \(5 = (\sqrt{5})^2\), every power of 5 is an even power of \(\sqrt{5}\). The first term is \((\sqrt{5})^0\), so the term \((\sqrt{5})^8\) is the 9th term, not the 8th.

Mark scheme

1 mark
M1 for writing 625 as a power of 5 or of \(\sqrt{5}\), e.g. \(625 = 5^4\)
or listing terms: 1, \(\sqrt{5}\), 5, \(5\sqrt{5}\), 25, ..; correct answer with no working scores full marks
1 mark
M1 for \(5^4 = (\sqrt{5})^8\)
or continuing the list to 625
1 mark
A1 for the 9th term
cao

Question 4

4 marks

A geometric sequence has first term \(2 + \sqrt{3}\) and second term 1

Find the common ratio, giving your answer in the form \(a + b\sqrt{3}\) where \(a\) and \(b\) are integers. Hence write down the third term of the sequence.

Model answer

\(r = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}\)
Third term \(= 1 \times (2 - \sqrt{3}) = 2 - \sqrt{3}\)

Explanation

The common ratio is the second term divided by the first. Rationalise the denominator using \((2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1\). Because the second term is 1, the third term equals the ratio itself.

Mark scheme

1 mark
M1 for \(r = \dfrac{1}{2 + \sqrt{3}}\)
correct answer with no working scores full marks
1 mark
M1 for multiplying the numerator and denominator by \(2 - \sqrt{3}\)
1 mark
A1 for \(r = 2 - \sqrt{3}\)
1 mark
B1 for a third term of \(2 - \sqrt{3}\)
ft their \(r\) (third term \(= 1 \times r\))

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