Geometric sequences and series - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 18 exam-style questions on Geometric sequences and series for Edexcel A-Level Maths (Paper 1). 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.

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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

Find the sum of the terms of the geometric series
\[6 + 9 + 13.5 + \ldots\]
from the 8th term to the 20th term inclusive. Give your answer correct to 3 significant figures.

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

Model answer

\(r = \dfrac{9}{6} = 1.5\)
Sum from the 8th to the 20th term \(= S_{20} - S_7\)
\(S_{20} = \dfrac{6(1.5^{20} - 1)}{1.5 - 1} = 39891.08\ldots\)
\(S_7 = \dfrac{6(1.5^{7} - 1)}{1.5 - 1} = 193.03\ldots\)
\(S_{20} - S_7 = 39698.05\ldots = 39700\) (3 s.f.)

Explanation

The terms from the 8th to the 20th are the first 20 terms minus the first 7, so the sum is \(S_{20} - S_7\). The method marks are for using the sum formula with \(r = 1.5\) and for subtracting the right sum. The most common error is subtracting \(S_8\), which removes the 8th term as well.

Mark scheme

1 mark
M1 Uses the sum formula with \(a = 6\) and \(r = 1.5\) for a sum of terms, e.g. \(S_{20} = \dfrac{6(1 - 1.5^{20})}{1 - 1.5}\) \((= 39891.08\ldots)\); or a correct final answer of awrt 39700 with no working
An answer of awrt 39700 with no working scores 3 marks. Alternative method: a sum of 13 terms with first term \(u_8 = 6 \times 1.5^7 = 102.515625\), i.e. \(\dfrac{102.515625(1.5^{13} - 1)}{1.5 - 1}\), scores M1 M1, then A1 for awrt 39700.
1 mark
M1 (dM1) A correct strategy for the 8th to the 20th terms: \(S_{20} - S_7\) (with \(S_7 = 193.03\ldots\)), or 13 terms starting from \(u_8\)
A common error is \(S_{20} - S_8\) (giving 39595.5...), which removes the 8th term as well and scores M1M0A0. \(S_{19} - S_7\) also scores M1M0A0. Accept the values of \(S_{20}\) and \(S_7\) to 3 s.f. or better.
1 mark
A1 awrt 39700 (3 s.f.), i.e. in [39650, 39750)
The value is 39698.05 to 2 d.p.; accept 39698 or 39698.0. Do not accept 39600 or 40000.

Question 2

5 marks

A geometric series has first term \(a\) and common ratio \(r\), where \(r ≠ 1\).

Prove that the sum of the first \(n\) terms of the series is
\[S_n = \dfrac{a(1 - r^n)}{1 - r}\]
Hence explain why the series has a sum to infinity of \(\dfrac{a}{1 - r}\) when \(\lvert r \rvert < 1\).

Model answer

\(S_n = a + ar + ar^2 + \ldots + ar^{n-2} + ar^{n-1}\)
\(rS_n = ar + ar^2 + \ldots + ar^{n-1} + ar^n\)
Subtracting: \(S_n - rS_n = a - ar^n\)
\(S_n(1 - r) = a(1 - r^n)\)
Since \(r ≠ 1\), \(S_n = \dfrac{a(1 - r^n)}{1 - r}\), as required.
If \(\lvert r \rvert < 1\), then \(r^n \to 0\) as \(n \to \infty\), so \(S_n \to \dfrac{a(1 - 0)}{1 - r} = \dfrac{a}{1 - r}\).
Hence the sum to infinity is \(\dfrac{a}{1 - r}\).

Explanation

Write out \(S_n\), multiply every term by \(r\), and subtract so that all the terms except \(a\) and \(ar^n\) cancel. The B mark needs the last term to be \(ar^{n-1}\), and the final mark needs every step, including dividing by \(1 - r\), which is allowed because \(r ≠ 1\). For the sum to infinity, the key point is that \(r^n \to 0\) when \(\lvert r \rvert < 1\).

Mark scheme

1 mark
B1 Writes \(S_n = a + ar + ar^2 + \ldots + ar^{n-1}\), with at least 3 correct terms including the first and the last, and no incorrect terms. An answer with no working scores 0 marks
An answer with no working scores 0 marks: restating the printed result earns nothing. "\(S_n =\)" must be seen (or implied by the next line). A last term of \(ar^n\) scores B0, but the M marks may still be earned.
1 mark
M1 Multiplies by \(r\): \(rS_n = ar + ar^2 + \ldots + ar^n\)
Allow for their series from the B mark. A proof using numerical values of \(a\) and \(r\) scores 0 marks.
1 mark
M1 (dM1) Subtracts the two series so that the middle terms cancel: \(S_n - rS_n = a - ar^n\) (or \(rS_n - S_n = ar^n - a\))
This is the key step. Writing the two series without subtracting them scores M0.
1 mark
A1* Factorises to \(S_n(1 - r) = a(1 - r^n)\) and divides by \(1 - r\) (allowed as \(r ≠ 1\)) to reach \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) (cso)
From \(rS_n - S_n = ar^n - a\), a correct step such as multiplying the numerator and denominator by \(-1\) must be seen to reach the printed form. There must be no errors in the proof, including invisible brackets.
1 mark
B1 States that \(r^n \to 0\) as \(n \to \infty\) because \(\lvert r \rvert < 1\), so \(S_n \to \dfrac{a(1 - 0)}{1 - r} = \dfrac{a}{1 - r}\)
Requires both the reason (\(\lvert r \rvert < 1\), so \(r^n\) tends to 0) and its effect on \(S_n\). Accept "\(r^n\) becomes negligible as \(n\) increases". Do not accept "\(r^n = 0\)", or "\(r^n\) gets smaller" without saying that it tends to 0.

Question 3

6 marks

An infinite geometric series is
\[2 + \dfrac{8}{x + 1} + \dfrac{32}{(x + 1)^2} + \ldots, \quad x ≠ -1\]
Find, using set notation, the set of values of \(x\) for which the series is convergent.

Given that the series has sum to infinity 6, find the value of \(x\).

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

Model answer

\(r = \dfrac{8}{x + 1} \div 2 = \dfrac{4}{x + 1}\)
The series converges when \(\lvert r \rvert < 1\): \(\left\lvert \dfrac{4}{x + 1} \right\rvert < 1 \Rightarrow \lvert x + 1 \rvert > 4\)
\(x + 1 > 4 \Rightarrow x > 3\), or \(x + 1 < -4 \Rightarrow x < -5\)
Set of values: \(\{x : x < -5\} \cup \{x : x > 3\}\)
\(S_\infty = \dfrac{2}{1 - \frac{4}{x + 1}} = \dfrac{2(x + 1)}{x - 3} = 6\)
\(2x + 2 = 6x - 18 \Rightarrow 4x = 20 \Rightarrow x = 5\)
Check: \(x = 5\) gives \(r = \dfrac{2}{3}\), so \(\lvert r \rvert < 1\). So \(x = 5\)

Explanation

The common ratio is \(\dfrac{4}{x + 1}\) and the series converges only when \(\lvert r \rvert < 1\), which gives \(\lvert x + 1 \rvert > 4\): two separate intervals, joined with ∪ in set notation. Setting \(\dfrac{a}{1 - r} = 6\) gives a linear equation with solution \(x = 5\), which lies in the set. Common errors are giving the inside region \(-5 < x < 3\) and using ∩ to join the intervals.

Mark scheme

1 mark
M1 Finds \(r = \dfrac{4}{x + 1}\) and uses \(\lvert r \rvert < 1\): \(\left\lvert \dfrac{4}{x + 1} \right\rvert < 1\), i.e. \(\lvert x + 1 \rvert > 4\), or \(-1 < \dfrac{4}{x + 1} < 1\); or correct final answers of \(\{x : x < -5\} \cup \{x : x > 3\}\) and \(x = 5\) with no working (special case: 2 marks)
Correct answers of \(\{x : x < -5\} \cup \{x : x > 3\}\) and \(x = 5\) with no working score 2 marks (special case: 1 mark for each). Allow \((x + 1)^2 > 16\).
1 mark
A1 Critical values \(x = -5\) and \(x = 3\)
Both are needed.
1 mark
A1 \(\{x : x < -5\} \cup \{x : x > 3\}\)
Requires { } and ∪. Accept \(\{x \in \mathbb{R} : x < -5\} \cup \{x \in \mathbb{R} : x > 3\}\) and typed forms such as {x : x < -5} U {x : x > 3}. Do not accept \(x < -5\) or \(x > 3\) without set notation, interval notation such as \((-\infty, -5) \cup (3, \infty)\) (not the set notation asked for), ∩, \(\{x : -5 < x < 3\}\), or inequalities that include \(-5\) or 3.
1 mark
M1 Uses \(S_\infty = \dfrac{2}{1 - \frac{4}{x + 1}} = 6\) with their \(r\) in terms of \(x\)
Independent of the first three marks.
1 mark
M1 (dM1) Clears the fractions and solves the resulting linear equation, e.g. \(2(x + 1) = 6(x - 3)\)
Allow slips in the rearrangement for this mark.
1 mark
A1 \(x = 5\)
\(x = 5\) gives \(r = \dfrac{2}{3}\), which satisfies \(\lvert r \rvert < 1\). A value outside their set should be rejected.

Question 4

9 marks

A drinks company makes cans from 800 kg of new aluminium. A model assumes that, when cans are recycled, 75% of their aluminium is recovered and used to make new cans, and the rest is lost. The new cans are recycled in the same way, and so on.

The cans made in round 1 contain 800 kg of aluminium, the cans made from them in round 2 contain 600 kg, and so on.

Using the model, find the total mass of aluminium in all the cans made in the first 10 rounds, giving your answer in kg to 3 significant figures.

Using logarithms, find the number of the first round in which the cans made contain less than 1 kg of aluminium.

Show that, however many rounds there are, the model predicts that the total mass of aluminium in all the cans made is less than 3.2 tonnes.

The company wants the model to predict that, in the long term, the total mass of aluminium in all the cans made from each 800 kg of new aluminium is at least 5 tonnes. Find the least percentage of the aluminium in recycled cans that must be recovered.

Give one limitation of the model.

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

Model answer

Masses in each round: 800, 600, 450, ... kg: geometric with \(a = 800\), \(r = 0.75\)
\(S_{10} = \dfrac{800(1 - 0.75^{10})}{1 - 0.75} = 3200(1 - 0.75^{10}) = 3019.8\ldots\)
Total in the first 10 rounds \(\approx 3020\) kg
Round \(n\) contains \(800 \times 0.75^{n - 1}\) kg.
\(800 \times 0.75^{n - 1} < 1\)
\(0.75^{n - 1} < \dfrac{1}{800}\)
\((n - 1)\ln 0.75 < \ln\dfrac{1}{800}\)
\(n - 1 > \dfrac{\ln(1/800)}{\ln 0.75} = 23.2\ldots\) (the inequality reverses as \(\ln 0.75 < 0\))
So \(n - 1 = 24\): the first such round is round 25.
\(S_n = 3200(1 - 0.75^n)\), and \(0.75^n > 0\), so \(S_n < 3200\) kg \(= 3.2\) tonnes for every \(n\). (\(S_\infty = \dfrac{800}{1 - 0.75} = 3200\) kg.)
Let \(p\) be the proportion recovered: \(S_\infty = \dfrac{800}{1 - p} \geqslant 5000\)
\(1 - p \leqslant 0.16\), so \(p \geqslant 0.84\): at least 84% must be recovered.
Limitation: in practice not all cans are returned for recycling, so less than 75% of the aluminium would be recovered in each round.

Explanation

The masses in successive rounds form a geometric sequence with first term 800 and ratio 0.75, so the total for 10 rounds is \(S_{10}\), and every finite total is below the sum to infinity, 3200 kg. For the round with less than 1 kg, the nth term is \(800 \times 0.75^{n - 1}\), and dividing by the negative \(\ln 0.75\) reverses the inequality; for a long-term total of 5000 kg the ratio must satisfy \(\dfrac{800}{1 - p} \geqslant 5000\). Common errors are using \(800 \times 0.75^n\) for the nth term (giving round 24) and mixing kilograms with tonnes.

Mark scheme

1 mark
M1 Uses \(S_{10} = \dfrac{800(1 - 0.75^{10})}{1 - 0.75}\) with \(a = 800\) and \(r = 0.75\); or correct final answers of 3020 kg and 84% with no working (special case: 1 mark each)
An answer of 3020 kg with no working scores 1 mark and an answer of 84% with no working scores 1 mark (special cases). An answer of round 25 with no working scores 0 marks, because the question demands logarithms; the 'show that' needs working, and the limitation earns its own B mark.
1 mark
A1 awrt 3020 kg, i.e. in [3015, 3025) (the exact value is 3019.8...)
Accept awrt 3.02 tonnes, i.e. in [3.015, 3.025). Using \(n = 9\) (2960 kg) or a ratio of 0.25 scores A0.
1 mark
M1 Sets \(800 \times 0.75^{n - 1} < 1\) (or \(= 1\)) and uses logarithms correctly: \(n - 1 > \dfrac{\log(1/800)}{\log 0.75}\) \((= 23.2\ldots)\)
Dividing by \(\log 0.75\), which is negative, reverses the inequality. Using \(800 \times 0.75^n\) (one round out) can score M1 but leads to A0. Trial of values without logarithms scores M0. Accept 23.2 or a more accurate value (23.236...).
1 mark
A1 Round 25
cso. Accept \(n = 25\). Do not accept 24 (the cans in round 24 contain 1.07 kg).
1 mark
M1 Uses \(S_\infty = \dfrac{800}{1 - 0.75} = 3200\) kg, or writes \(S_n = 3200(1 - 0.75^n)\)
1 mark
A1* Concludes that \(S_n = 3200(1 - 0.75^n) < 3200\) kg \(= 3.2\) tonnes for every \(n\), because \(0.75^n > 0\) (or because every term is positive, so the total of any number of rounds is less than the sum to infinity), cso
A conclusion with a reason is needed; writing \(S_\infty = 3200\) with no comparison scores A0. Mixing kg and tonnes without converting scores A0.
1 mark
M1 Sets \(\dfrac{800}{1 - p} \geqslant 5000\) (or \(= 5000\)), where \(p\) is the proportion recovered, and solves for \(p\)
Using 5 with 800 (inconsistent units) scores M0.
1 mark
A1 84%
Accept \(p = 0.84\), or 'at least 84%'.
1 mark
B1 A limitation in context, e.g. not all cans are returned for recycling; the proportion recovered may not be 75% (or the same) in every round; the model allows infinitely many rounds, but in practice the aluminium cannot be recycled indefinitely or in tiny amounts; aluminium is also lost in other ways
Do not accept vague answers such as "it is only a model" or "it is inaccurate".

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