Geometric sequences and series - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Geometric sequences and series for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

18 questions
2–10 marks each
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18 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

3 marks

A quiz has 8 prizes. The first prize is £\(P\), and each prize after the first is worth 75% of the prize before it. The total value of the 8 prizes is £10 000.

Find the value of \(P\), correct to the nearest penny.

[Geometric series: \(S_n = \dfrac{a(1 - r^n)}{1 - r}\)]

Model answer

The prizes form a geometric series with first term \(P\) and common ratio 0.75.
\(S_8 = \dfrac{P(1 - 0.75^8)}{1 - 0.75} = 10000\)
\(4P(1 - 0.100112\ldots) = 10000\)
\(3.59955\ldots P = 10000\)
\(P = 2778.126\ldots\)
The first prize is £2778.13 (to the nearest penny).

Explanation

The prizes form a geometric sequence with first term \(P\) and common ratio 0.75, so their total is \(S_8\). Setting \(S_8 = 10000\) and solving gives \(P\), which must be rounded to the nearest penny. A common error is to use the 8th term, \(P \times 0.75^7\), instead of the sum.

Mark scheme

1 mark
M1 Uses the sum formula with \(r = 0.75\) and \(n = 8\) and sets it equal to 10 000: \(\dfrac{P(1 - 0.75^8)}{1 - 0.75} = 10000\); or a correct final answer of £2778.13 with no working
An answer of £2778.13 with no working scores 3 marks. Using \(n = 7\), or setting the 8th prize \(P \times 0.75^7\) equal to 10 000, scores M0.
1 mark
A1 A correct equation with the power evaluated, e.g. \(4P(1 - 0.1001\ldots) = 10000\) or \(3.5995\ldots P = 10000\)
Accept \(P = \dfrac{2500}{1 - 0.75^8}\) oe.
1 mark
A1 \(P = 2778.13\)
Depends on the M mark. Must be to the nearest penny: do not accept 2778.1, 2778 or 2780 as the final answer. Units not required. The exact value is 2778.126...

Question 2

6 marks

A geometric sequence begins \(u_1, u_2, u_3, \ldots\), where
\[u_1 + u_2 = 12 \quad \text{and} \quad u_2 + u_3 = -6\]
Find the common ratio and the first term of the sequence.

Explain why the corresponding geometric series has a sum to infinity, and find its value.

[Geometric series: \(S_\infty = \dfrac{a}{1 - r}\) for \(\lvert r \rvert < 1\)]

Model answer

\(a + ar = 12\) and \(ar + ar^2 = -6\)
\(ar + ar^2 = r(a + ar)\), so \(12r = -6\) and \(r = -\dfrac{1}{2}\)
\(a\left(1 - \dfrac{1}{2}\right) = 12\), so \(a = 24\)
Since \(\lvert r \rvert = \dfrac{1}{2} < 1\), the series is convergent and has a sum to infinity.
\(S_\infty = \dfrac{24}{1 - \left(-\frac{1}{2}\right)} = \dfrac{24}{1.5} = 16\)

Explanation

Each of the pairs \(u_2 + u_3\) and \(u_1 + u_2\) is a multiple of the other: \(u_2 + u_3 = r(u_1 + u_2)\), so \(r = -\frac{1}{2}\) and then \(a = 24\). The sum to infinity exists because \(\lvert r \rvert < 1\), and \(\frac{a}{1 - r} = 16\). A common error is the sign of \(r\) in \(1 - r\), giving \(\frac{24}{0.5} = 48\).

Mark scheme

1 mark
M1 Writes the conditions as \(a + ar = 12\) and \(ar + ar^2 = -6\) and eliminates \(a\), e.g. \(\dfrac{ar(1 + r)}{a(1 + r)} = r = \dfrac{-6}{12}\); or correct final answers of \(r = -\tfrac{1}{2}\), \(a = 24\) and 16 with no working
Correct answers of \(r = -\tfrac{1}{2}\), \(a = 24\) and a sum to infinity of 16 with no working score 5 marks (the reason still needs to be given for the B mark). Alternative: substitutes \(a = \dfrac{12}{1 + r}\) into \(ar + ar^2 = -6\) to reach \(12r = -6\).
1 mark
A1 \(r = -\dfrac{1}{2}\)
Accept \(-0.5\).
1 mark
A1 \(a = 24\)
Depends on the M mark. From \(a\left(1 - \tfrac{1}{2}\right) = 12\).
1 mark
B1 The sum to infinity exists because \(\lvert r \rvert = \dfrac{1}{2} < 1\)
B1ft on their \(r\) if \(\lvert r \rvert < 1\). Requires reference to \(\lvert r \rvert < 1\) (or \(-1 < r < 1\)); "the terms get smaller" alone scores B0.
1 mark
M1 Uses \(S_\infty = \dfrac{a}{1 - r}\) with their \(a\) and \(r\): \(\dfrac{24}{1 - \left(-\frac{1}{2}\right)}\)
Using \(\dfrac{a}{1 + r}\) scores M0.
1 mark
A1 \(S_\infty = 16\)
cao

Question 3

6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The infinite series
\[\mathrm{e}^{-x} + \mathrm{e}^{-3x} + \mathrm{e}^{-5x} + \ldots, \quad x > 0\]
is geometric.

Explain why the series is convergent for all \(x > 0\), and show that \(S_\infty\), its sum to infinity, is equal to \(\dfrac{1}{\mathrm{e}^{x} - \mathrm{e}^{-x}}\).

Given that \(S_\infty = \dfrac{3}{8}\), find the exact value of \(x\).

[Geometric series: \(S_\infty = \dfrac{a}{1 - r}\) for \(\lvert r \rvert < 1\)]

Model answer

The common ratio is \(r = \dfrac{\mathrm{e}^{-3x}}{\mathrm{e}^{-x}} = \mathrm{e}^{-2x}\)
For \(x > 0\), \(-2x < 0\), so \(0 < \mathrm{e}^{-2x} < 1\). Hence \(\lvert r \rvert < 1\) and the series is convergent.
\(S_\infty = \dfrac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-2x}}\)
Multiplying the numerator and the denominator by \(\mathrm{e}^{x}\): \(S_\infty = \dfrac{1}{\mathrm{e}^{x} - \mathrm{e}^{-x}}\), as required
\(\dfrac{1}{\mathrm{e}^{x} - \mathrm{e}^{-x}} = \dfrac{3}{8}\)
\(3\mathrm{e}^{x} - 3\mathrm{e}^{-x} = 8\)
\(3\mathrm{e}^{2x} - 8\mathrm{e}^{x} - 3 = 0\)
\((3\mathrm{e}^{x} + 1)(\mathrm{e}^{x} - 3) = 0\)
\(\mathrm{e}^{x} = -\dfrac{1}{3}\) is impossible, as \(\mathrm{e}^{x} > 0\), so \(\mathrm{e}^{x} = 3\)
\(x = \ln 3\)

Explanation

Each term is the one before multiplied by \(\mathrm{e}^{-2x}\), which lies between 0 and 1 when \(x > 0\), so the sum to infinity exists. Multiplying the numerator and the denominator of \(\frac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-2x}}\) by \(\mathrm{e}^{x}\) gives the printed form, and the equation then becomes a quadratic in \(\mathrm{e}^{x}\). A common error is to keep \(\mathrm{e}^{x} = -\frac{1}{3}\), which has no solution.

Mark scheme

1 mark
B1 States that the common ratio is \(r = \mathrm{e}^{-2x}\) and that \(0 < \mathrm{e}^{-2x} < 1\) when \(x > 0\), so \(\lvert r \rvert < 1\) and the series is convergent. An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question demands working, and \(x = \ln 3\) with no working scores 0 marks. A reason why \(\mathrm{e}^{-2x} < 1\) is needed, e.g. \(-2x < 0\) so \(\mathrm{e}^{-2x} < \mathrm{e}^0 = 1\). Do not accept "the terms get smaller" alone.
1 mark
M1 Uses \(S_\infty = \dfrac{a}{1 - r}\) with \(a = \mathrm{e}^{-x}\) and \(r = \mathrm{e}^{-2x}\): \(S_\infty = \dfrac{\mathrm{e}^{-x}}{1 - \mathrm{e}^{-2x}}\)
1 mark
A1* Multiplies the numerator and the denominator by \(\mathrm{e}^{x}\) to reach \(S_\infty = \dfrac{1}{\mathrm{e}^{x} - \mathrm{e}^{-x}}\) cso
The step \(\mathrm{e}^{x} \times \mathrm{e}^{-2x} = \mathrm{e}^{-x}\) must be seen or clearly implied.
1 mark
M1 Sets \(\dfrac{1}{\mathrm{e}^{x} - \mathrm{e}^{-x}} = \dfrac{3}{8}\) and multiplies through by \(\mathrm{e}^{x}\) to form a 3-term quadratic in \(\mathrm{e}^{x}\): \(3\mathrm{e}^{2x} - 8\mathrm{e}^{x} - 3 = 0\)
Independent of the earlier marks: may use the printed result. Allow \(u = \mathrm{e}^{x}\), \(3u^2 - 8u - 3 = 0\). Condone a slip in collecting terms, but it must be a 3-term quadratic in \(\mathrm{e}^{x}\).
1 mark
M1 (dM1) Solves their quadratic by factorising, the formula or completing the square: \((3\mathrm{e}^{x} + 1)(\mathrm{e}^{x} - 3) = 0\)
Depends on the previous M mark. Roots written down from a calculator with no factorising, formula or completing the square seen score M0, as the question demands working.
1 mark
A1 \(x = \ln 3\) only, rejecting \(\mathrm{e}^{x} = -\dfrac{1}{3}\) because \(\mathrm{e}^{x} > 0\)
A second answer such as \(\ln\left(-\tfrac{1}{3}\right)\) scores A0. A decimal only (1.0986...) scores A0 as the exact value is asked for.

Question 4

8 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The infinite series
\[1 + 2\sin x\cos x + 4\sin^2 x\cos^2 x + 8\sin^3 x\cos^3 x + \ldots, \quad 0 < x < \pi\]
is geometric. Where it is convergent, its sum to infinity is \(S_\infty\).

Find the values of \(x\) in the interval \(0 < x < \pi\) for which the series is not convergent.

Find the range of possible values of \(S_\infty\).

Find the set of values of \(x\) for which \(S_\infty > 2\). Give your answer in set notation.

[Geometric series: \(S_\infty = \dfrac{a}{1 - r}\) for \(\lvert r \rvert < 1\)]

Model answer

The common ratio is \(r = 2\sin x\cos x = \sin 2x\): each term is \(\sin 2x\) times the term before.
The series converges when \(\lvert \sin 2x \rvert < 1\). Since \(\lvert \sin 2x \rvert \leqslant 1\) always, it is not convergent only when \(\sin 2x = \pm 1\).
For \(0 < x < \pi\), \(0 < 2x < 2\pi\): \(\sin 2x = 1\) gives \(2x = \dfrac{\pi}{2}\), and \(\sin 2x = -1\) gives \(2x = \dfrac{3\pi}{2}\).
The series is not convergent when \(x = \dfrac{\pi}{4}\) or \(x = \dfrac{3\pi}{4}\).
Otherwise \(S_\infty = \dfrac{1}{1 - \sin 2x}\), where \(-1 < \sin 2x < 1\), so \(0 < 1 - \sin 2x < 2\) and \(S_\infty > \dfrac{1}{2}\).
(\(\sin 2x\) takes every value between \(-1\) and 1, so \(S_\infty\) takes every value greater than \(\dfrac{1}{2}\).)
\(S_\infty > 2\): \(\dfrac{1}{1 - \sin 2x} > 2\), and \(1 - \sin 2x > 0\), so \(1 - \sin 2x < \dfrac{1}{2}\), i.e. \(\sin 2x > \dfrac{1}{2}\)
\(\sin 2x = \dfrac{1}{2}\) for \(0 < 2x < 2\pi\): \(2x = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\), so \(x = \dfrac{\pi}{12}\) or \(\dfrac{5\pi}{12}\)
\(\sin 2x > \dfrac{1}{2}\) when \(\dfrac{\pi}{12} < x < \dfrac{5\pi}{12}\), but the series is not convergent at \(x = \dfrac{\pi}{4}\).
Set of values: \(\left\{x : \dfrac{\pi}{12} < x < \dfrac{\pi}{4}\right\} \cup \left\{x : \dfrac{\pi}{4} < x < \dfrac{5\pi}{12}\right\}\)

Explanation

The double-angle formula shows that the common ratio is \(\sin 2x\), so the series converges except where \(\sin 2x = \pm 1\), at \(x = \frac{\pi}{4}\) and \(\frac{3\pi}{4}\). Because \(1 - \sin 2x\) lies between 0 and 2, the sum to infinity is always greater than \(\frac{1}{2}\), and \(S_\infty > 2\) reduces to \(\sin 2x > \frac{1}{2}\). A common error is to forget that \(x = \frac{\pi}{4}\), which lies inside the interval, must be removed because the series does not converge there.

Mark scheme

1 mark
B1 Uses \(\sin 2x \equiv 2\sin x\cos x\) to identify the common ratio as \(r = \sin 2x\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question demands working, so \(x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}\), \(S_\infty > \tfrac{1}{2}\) and the set of values with no working score 0 marks. Accept \(r = 2\sin x\cos x\) if it is then used correctly as \(\sin 2x\).
1 mark
M1 Uses the condition \(\lvert r \rvert < 1\): since \(\lvert \sin 2x \rvert \leqslant 1\) always, the series is not convergent only when \(\sin 2x = 1\) or \(\sin 2x = -1\)
Solving \(\lvert 2\sin x\cos x \rvert = 1\) is equivalent.
1 mark
A1 \(x = \dfrac{\pi}{4}\) and \(x = \dfrac{3\pi}{4}\) only
Both values, exact and in radians, and no others in \(0 < x < \pi\). Answers in degrees (45° and 135°) score A0.
1 mark
B1 \(S_\infty > \dfrac{1}{2}\), from \(S_\infty = \dfrac{1}{1 - \sin 2x}\) with \(-1 < \sin 2x < 1\), so that \(0 < 1 - \sin 2x < 2\)
Accept \(\left\{S_\infty : S_\infty > \tfrac{1}{2}\right\}\) or \(S_\infty \in \left(\tfrac{1}{2}, \infty\right)\). \(S_\infty \geqslant \tfrac{1}{2}\), or an upper limit for \(S_\infty\), scores B0.
1 mark
M1 Sets \(\dfrac{1}{1 - \sin 2x} > 2\) and, since \(1 - \sin 2x > 0\), rearranges to \(\sin 2x > \dfrac{1}{2}\)
Independent of the earlier marks. Allow an equation for this mark if the inequality is used correctly later.
1 mark
M1 (dM1) Solves \(\sin 2x = \dfrac{1}{2}\) for \(0 < 2x < 2\pi\): \(2x = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\), so \(x = \dfrac{\pi}{12}\) or \(\dfrac{5\pi}{12}\)
Depends on the previous M mark. Both critical values are needed; values in degrees (15° and 75°) can score this mark.
1 mark
A1 \(\dfrac{\pi}{12} < x < \dfrac{5\pi}{12}\) for \(\sin 2x > \dfrac{1}{2}\)
The inside region, seen in any form (before or after \(\tfrac{\pi}{4}\) is removed), exact and in radians.
1 mark
A1 \(\left\{x : \dfrac{\pi}{12} < x < \dfrac{\pi}{4}\right\} \cup \left\{x : \dfrac{\pi}{4} < x < \dfrac{5\pi}{12}\right\}\)
Depends on both of the last two M marks. Requires \(x = \tfrac{\pi}{4}\) removed (the series is not convergent there) and set notation; accept \(\left\{x : \tfrac{\pi}{12} < x < \tfrac{5\pi}{12}, \ x ≠ \tfrac{\pi}{4}\right\}\). Including \(\tfrac{\pi}{4}\), \(\leqslant\) signs, or no set notation scores A0.

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