P4.5Geometric sequences and series

Edexcel A level Maths (9MA0) · Pure mathematics › Sequences and series

Practise Geometric sequences and series. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy6 marks

A geometric series has first term \(126\) and common ratio \(-\frac{3}{4}\).

(a) Find the 5th term of the series, giving your answer to 3 significant figures.[2]

(b) Find the sum of the first 10 terms of the series, giving your answer to 3 significant figures.[2]

Show the answer and mark scheme

(a) Answer: \(39.9\)

M1 for \(126 \times \left(-\frac{3}{4}\right)^{4}\)
A1 for awrt \(39.9\)

Worked solution: \(u_{5} = ar^{4} = 126 \times \left(-\frac{3}{4}\right)^{4} = 39.86718\ldots = 39.9\)

(b) Answer: \(67.9\)

M1 for \(\frac{126\left(1 - \left(-\frac{3}{4}\right)^{10}\right)}{1 - \left(-\frac{3}{4}\right)}\)
A1 for awrt \(67.9\)

Worked solution: \(S_{10} = \frac{126\left(1 - \left(-\frac{3}{4}\right)^{10}\right)}{1 - \left(-\frac{3}{4}\right)} = 67.9454\ldots = 67.9\)

M1 for \(\frac{126}{1 - \left(-\frac{3}{4}\right)}\)
A1 for \(72\)

Worked solution: \(|r| \lt 1\), so \(S_\infty = \frac{a}{1 - r} = \frac{126}{1 - \left(-\frac{3}{4}\right)} = 72\).

Question 2Medium7 marks

A geometric series has first term 3 and common ratio 1.2.

(a) Show that the sum of the first \(n\) terms of the series exceeds 1000 when \(1.2^n \gt \dfrac{203}{3}\).[2]

(b) Hence find the smallest value of \(n\) for which the sum of the first \(n\) terms exceeds 1000.[2]

(c) A second geometric series has first term 20 and common ratio 0.8. A student wants the smallest \(n\) for which the sum of the first \(n\) terms, \(S_n\), is within 1 of the sum to infinity, \(S_\infty\). The student writes:
\(S_\infty - S_n = 100 \times 0.8^n \lt 1\)
\(n \log 0.8 \lt \log 0.01\)
\(n \lt \frac{\log 0.01}{\log 0.8} = 20.6\), so \(n = 20\).
Explain the error in the student's working and find the correct value of \(n\).[3]

Show the answer and mark scheme

(a) Answer: \(S_n = \frac{3(1.2^n - 1)}{0.2} = 15(1.2^n - 1) \gt 1000 \Rightarrow 1.2^n \gt 1 + \frac{200}{3} = \frac{203}{3}\)

M1 for \(S_n = \frac{3(1.2^n - 1)}{1.2 - 1}\)
A1* for \(15(1.2^n - 1) \gt 1000 \Rightarrow 1.2^n \gt \frac{203}{3}\)

Worked solution: \(S_n = \dfrac{3(1.2^n - 1)}{1.2 - 1} = 15(1.2^n - 1)\). \(S_n \gt 1000 \iff 1.2^n - 1 \gt \frac{200}{3} \iff 1.2^n \gt \frac{203}{3}\).

(b) Answer: 24

M1 for \(n \gt \frac{\log(203/3)}{\log 1.2}\) (= 23.1…)
A1 for 24

Worked solution: \(n \log 1.2 \gt \log \frac{203}{3}\), and \(\log 1.2 \gt 0\), so \(n \gt \dfrac{\log(203/3)}{\log 1.2} = 23.1\ldots\). The smallest integer is \(n = 24\) (check: \(S_{23} = 978.7\), \(S_{24} = 1177.5\)).

B1 for \(\log 0.8 \lt 0\), so dividing by it reverses the inequality
M1 for \(n \gt 20.6\)
A1 for 21

Worked solution: \(S_\infty = \frac{20}{1 - 0.8} = 100\) and \(S_\infty - S_n = 100 \times 0.8^n\), so the first two lines are correct. But \(\log 0.8\) is negative, and dividing an inequality by a negative number reverses it: \(n \gt \dfrac{\log 0.01}{\log 0.8} = 20.6\ldots\).
So the smallest value is \(n = 21\) (check: \(100 \times 0.8^{20} = 1.15\), \(100 \times 0.8^{21} = 0.92\)).

Question 3Hard10 marks

An infinite geometric series has first term \(a\) and common ratio \(r\), where \(-1 \lt r \lt 1\) and \(r \ne 0\). Its sum to infinity is \(S\).
The squares of the terms of this series form a second series, whose sum to infinity is \(T\).

(a) Explain why the second series is geometric and converges, and show that \(T = \dfrac{a^2}{1 - r^2}\).[2]

(b) Given that \(S = 12\) and \(T = 48\), find the values of \(a\) and \(r\).[4]

(d) Find the first term and common ratio of an infinite geometric series with \(S = 1\) and \(T = 4\).[3]

Show the answer and mark scheme

(a) Answer: The squares are \(a^2, a^2r^2, a^2r^4, \ldots\): geometric with ratio \(r^2\), and \(0 \lt r^2 \lt 1\), so it converges to \(\frac{a^2}{1 - r^2}\).

M1 for the terms \(a^2, a^2r^2, a^2r^4, \ldots\) with common ratio \(r^2\), and \(0 \lt r^2 \lt 1\) so it converges
A1* for \(T = \frac{a^2}{1 - r^2}\)

Worked solution: The terms of the first series are \(ar^{k - 1}\), so their squares are \(a^2r^{2(k - 1)} = a^2(r^2)^{k - 1}\): a geometric series with ratio \(r^2\). As \(0 \lt r^2 \lt 1\) it converges, with sum \(T = \dfrac{a^2}{1 - r^2}\).

(b) Answer: \(a = 6\), \(r = \frac{1}{2}\)

M1 for \(\frac{a}{1 - r} = 12\) and \(\frac{a^2}{1 - r^2} = 48\)
M1 for using \(1 - r^2 = (1 - r)(1 + r)\) to obtain \(\frac{a}{1 + r} = 4\)
M1 for solving \(12(1 - r) = 4(1 + r)\) (or equivalent)
A1 for \(r = \frac{1}{2}\) and \(a = 6\)

Worked solution: \(\dfrac{a^2}{(1 - r)(1 + r)} = 48\) and \(\dfrac{a}{1 - r} = 12\). Dividing: \(\dfrac{a}{1 + r} = 4\).
So \(a = 12(1 - r) = 4(1 + r)\), giving \(12 - 12r = 4 + 4r\), \(r = \frac{1}{2}\) and \(a = 6\).

B1* for \(\frac{T}{S^2} = \frac{a^2}{(1 - r)(1 + r)} \times \frac{(1 - r)^2}{a^2} = \frac{1 - r}{1 + r}\)

Worked solution: \(\dfrac{T}{S^2} = \dfrac{a^2}{(1 - r)(1 + r)} \times \dfrac{(1 - r)^2}{a^2} = \dfrac{1 - r}{1 + r}\).

(d) Answer: \(r = -\frac{3}{5}\), \(a = \frac{8}{5}\)

M1 for \(\frac{1 - r}{1 + r} = 4\) (from part (c))
A1 for \(r = -\frac{3}{5}\) (which satisfies \(-1 \lt r \lt 1\))
A1 for \(a = S(1 - r) = \frac{8}{5}\)

Worked solution: \(\dfrac{1 - r}{1 + r} = \dfrac{4}{1^2} = 4 \Rightarrow 1 - r = 4 + 4r \Rightarrow r = -\frac{3}{5}\), which lies between \(-1\) and \(1\).
\(a = S(1 - r) = 1 \times \frac{8}{5} = \frac{8}{5}\). (Check: \(T = \dfrac{64/25}{16/25} = 4\).)

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