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

Practise 16 exam-style questions on Arithmetic 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.

16 questions
2–10 marks each
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16 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

The series
\[\sqrt{3} + \sqrt{12} + \sqrt{27} + \dots\]
is arithmetic.

Find the exact total of the terms of this series up to and including its 20th term, giving your answer in the form \(k\sqrt{3}\), where \(k\) is an integer.

[Arithmetic series: \(S_n = \dfrac{1}{2}n(a + l) = \dfrac{1}{2}n[2a + (n - 1)d]\)]

Model answer

\(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{27} = 3\sqrt{3}\), so the series is \(\sqrt{3} + 2\sqrt{3} + 3\sqrt{3} + \dots\) with \(a = \sqrt{3}\) and \(d = \sqrt{3}\)
\(S_{20} = \dfrac{20}{2}[2\sqrt{3} + 19\sqrt{3}] = 10 \times 21\sqrt{3} = 210\sqrt{3}\)

Explanation

Simplifying the surds shows that each term is a whole-number multiple of \(\sqrt{3}\), so the first term and the common difference are both \(\sqrt{3}\). The sum of 20 terms is \(10 \times 21\sqrt{3} = 210\sqrt{3}\), which is the same as \(\sqrt{3}(1 + 2 + \dots + 20)\). A common error is to take \(d = \sqrt{12} - \sqrt{3} = \sqrt{9} = 3\).

Mark scheme

1 mark
B1 Simplifies the terms to \(\sqrt{3}\), \(2\sqrt{3}\), \(3\sqrt{3}\), so \(a = \sqrt{3}\) and \(d = \sqrt{3}\); or a correct final answer of \(210\sqrt{3}\) with no working
An answer of \(210\sqrt{3}\) with no working scores 3 marks.
1 mark
M1 Uses the sum formula with \(n = 20\) and their \(a\) and \(d\): \(\dfrac{20}{2}[2\sqrt{3} + 19\sqrt{3}]\)
A correct formula with their values.
1 mark
A1 \(210\sqrt{3}\)
cao. A decimal (363.7) is A0.

Question 2

4 marks

In an arithmetic sequence, the sum of the first five terms is equal to the sum of the 6th, 7th and 8th terms. The 10th term of the sequence is 39.

Find the first term and the common difference of the sequence.

Model answer

Sum of the first five terms: \(\dfrac{5}{2}[2a + 4d] = 5a + 10d\)
Sum of the 6th, 7th and 8th terms: \((a + 5d) + (a + 6d) + (a + 7d) = 3a + 18d\)
\(5a + 10d = 3a + 18d\), so \(2a = 8d\) and \(a = 4d\)
10th term: \(a + 9d = 39\), so \(4d + 9d = 39\), \(13d = 39\), \(d = 3\)
\(a = 4d = 12\)
First term 12, common difference 3.

Explanation

Write every term as \(a + (n - 1)d\): the 6th, 7th and 8th terms are \(a + 5d\), \(a + 6d\) and \(a + 7d\), and the first five terms add to \(5a + 10d\). Equating gives \(a = 4d\), and the 10th term \(a + 9d = 39\) then gives \(d = 3\) and \(a = 12\). The commonest error is writing the \(n\)th term as \(a + nd\).

Mark scheme

1 mark
M1 Writes the sum of the first five terms as \(5a + 10d\) (or \(\dfrac{5}{2}[2a + 4d]\)) and the sum of the 6th, 7th and 8th terms as \(3a + 18d\), and equates them; or correct final answers of \(a = 12\) and \(d = 3\) with no working
Answers of \(a = 12\) and \(d = 3\) with no working score 4 marks. Using \(a + 6d\), \(a + 7d\), \(a + 8d\) for the 6th to 8th terms is M0.
1 mark
M1 (dM1) Simplifies to a relation such as \(a = 4d\) and uses \(a + 9d = 39\) to solve
Depends on the first M mark. \(a + 10d = 39\) is M0.
1 mark
A1 \(d = 3\)
cao.
1 mark
A1 \(a = 12\)
Depends on the second M mark. cao.

Question 3

6 marks

An arithmetic series has first term \(a\) and common difference \(d\). The sum of the first \(n\) terms of the series is \(S_n\).

Prove that
\[S_n = \dfrac{1}{2}n[2a + (n - 1)d]\]

Hence prove that the sum of any 12 consecutive positive integers is a multiple of 6 but is not a multiple of 12.

Model answer

\(S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d]\)
\(S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a\)
Adding term by term, each of the \(n\) pairs adds to \(2a + (n - 1)d\):
\(2S_n = n[2a + (n - 1)d]\), so \(S_n = \dfrac{1}{2}n[2a + (n - 1)d]\)
Let the 12 consecutive positive integers be \(m, m + 1, \dots, m + 11\): an arithmetic series with \(a = m\), \(d = 1\) and \(n = 12\).
\(S_{12} = \dfrac{12}{2}[2m + 11(1)] = 6(2m + 11)\)
\(2m + 11\) is an integer, so the sum is a multiple of 6.
\(2m\) is even, so \(2m + 11\) is odd, and 6 times an odd number is not a multiple of 12: \(6(2m + 11) = 12(m + 5) + 6\) leaves remainder 6 when divided by 12.
So the sum of any 12 consecutive positive integers is a multiple of 6 but is not a multiple of 12.

Explanation

Writing the series forwards and backwards and adding pairs the terms so that each pair is \(2a + (n - 1)d\), which gives \(2S_n = n[2a + (n - 1)d]\). Twelve consecutive integers form an arithmetic series with \(d = 1\), so their sum is \(6(2m + 11)\), and since \(2m + 11\) is odd the sum is a multiple of 6 but never of 12. The marks need the reversing step, a general first term \(m\) and a reason why \(2m + 11\) is odd; numerical checks are not proofs.

Mark scheme

1 mark
B1 Writes \(S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d]\), with at least three correct terms including the first and last terms and no incorrect terms. An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof. Using \(S_n = \frac{1}{2}n(a + l)\) without proving it scores 0 for the first three marks.
1 mark
M1 Reverses the series, \(S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + a\), and adds the two series term by term
The key step of reversing and adding must be seen.
1 mark
A1* \(2S_n = n[2a + (n - 1)d]\), so \(S_n = \dfrac{1}{2}n[2a + (n - 1)d]\)
cso. Must state that each of the \(n\) pairs adds to \(2a + (n - 1)d\), with \(S_n =\) seen.
1 mark
M1 Applies the formula to 12 consecutive integers starting at \(m\) (\(a = m\), \(d = 1\), \(n = 12\)): \(S_{12} = \dfrac{12}{2}[2m + 11] = 6(2m + 11)\) (or \(12m + 66\))
The given formula may be used even if it was not proved. Checking one or more numerical examples (e.g. \(1 + 2 + \dots + 12 = 78\)) is not a proof and scores 0 for these marks.
1 mark
A1 \(6(2m + 11)\), where \(2m + 11\) is an integer, so the sum is a multiple of 6
1 mark
A1 \(2m + 11\) is odd (\(2m\) is even), so \(6(2m + 11)\) is 6 times an odd number and is not a multiple of 12 (equivalently \(12m + 66 = 12(m + 5) + 6\) leaves remainder 6 when divided by 12), with a conclusion
Depends on the second M mark. cso. A reason why \(2m + 11\) is odd (or why the remainder is 6) and a (minimal) conclusion are needed.

Question 4

8 marks

Dev mows a large field in 26 parallel strips. The field is a trapezium, so each strip is longer than the one before. A model assumes that he mows the first strip in 10 minutes, and that each strip after that takes him 30 seconds longer than the strip before.

Show that, according to the model, the time in minutes that he takes to mow the first \(n\) strips is \(\dfrac{n(n + 39)}{4}\).

Find the total time that he takes to mow all 26 strips, in hours, minutes and seconds.

He starts at 9 am. Find the number of complete strips that he has mown by 1 pm.

He is paid a bonus if he mows the whole field in under 7 hours. Using the model, determine whether he earns the bonus.

Give one reason why the model may not be realistic.

[Arithmetic series: \(S_n = \dfrac{1}{2}n(a + l) = \dfrac{1}{2}n[2a + (n - 1)d]\)]

Model answer

The times for the strips (in minutes) form an arithmetic sequence with \(a = 10\) and \(d = 0.5\).
\(S_n = \dfrac{1}{2}n[2(10) + (n - 1)(0.5)] = \dfrac{1}{2}n(0.5n + 19.5) = \dfrac{n(n + 39)}{4}\)
26 strips: \(\dfrac{26 \times 65}{4} = 422.5\) minutes = 7 hours 2 minutes 30 seconds
By 1 pm he has mown for 240 minutes: \(\dfrac{n(n + 39)}{4} = 240 \Rightarrow n^2 + 39n - 960 = 0 \Rightarrow n = \dfrac{-39 + \sqrt{5361}}{2} = 17.1...\)
Check: \(S_{17} = \dfrac{17 \times 56}{4} = 238\) minutes, \(S_{18} = \dfrac{18 \times 57}{4} = 256.5\) minutes
So he has mown 17 complete strips by 1 pm.
422.5 minutes is more than 420 minutes (7 hours), so he does not earn the bonus.
The model may not be realistic because he is unlikely to slow down by exactly 30 seconds on every strip, and he may stop for breaks or to empty the grass box.

Explanation

The strip times form an arithmetic sequence with first term 10 minutes and common difference 0.5 minutes, so the time for \(n\) strips is the arithmetic series \(\dfrac{n(n + 39)}{4}\). The whole field takes 422.5 minutes, just over 7 hours, and solving \(\dfrac{n(n + 39)}{4} = 240\) shows that 17 complete strips are mown in the first 4 hours. Common errors are mixing seconds and minutes (using \(d = 30\) with \(a = 10\)) and using 4 instead of 240 minutes.

Mark scheme

1 mark
M1 Uses \(S_n = \dfrac{1}{2}n[2a + (n - 1)d]\) with \(a = 10\) and \(d = 0.5\) (minutes), or 600 and 30 (seconds). An answer with no working scores 0 marks for the show-that.
Answers of 7 h 2 min 30 s, 17 strips and "no" with no working score 3 marks (the B1 for the time, the final A1 for 17 strips and the B1 for the decision); the show-that needs working and the reason needs words.
1 mark
A1* \(\dfrac{1}{2}n[20 + 0.5(n - 1)] = \dfrac{1}{2}n(0.5n + 19.5) = \dfrac{n(n + 39)}{4}\)
cso, with an intermediate line. Working in seconds must be converted back to minutes.
1 mark
B1 \(\dfrac{26 \times 65}{4} = 422.5\) minutes = 7 hours 2 minutes 30 seconds
Both the minutes value (or equivalent) and the conversion needed. Accept 7 h 2.5 min.
1 mark
M1 Forms \(\dfrac{n(n + 39)}{4} = 240\) (4 hours = 240 minutes), i.e. \(n^2 + 39n - 960 = 0\), and solves; or evaluates the times for 17 and 18 strips
The given formula may be used even if it was not shown. Using 4 instead of 240 (mixed units) is M0.
1 mark
A1 \(n = 17.1...\), or the times 238 minutes for 17 strips and 256.5 minutes for 18 strips
The positive root is \(\dfrac{-39 + \sqrt{5361}}{2} = 17.11...\); accept awrt 17.1, i.e. [17.05, 17.15).
1 mark
A1 17 strips
Depends on the second M mark. 18 strips is A0.
1 mark
B1 No: 422.5 minutes is more than 7 hours (420 minutes), so he does not earn the bonus
Follow through their total time. A comparison of the two times is needed.
1 mark
B1 A sensible reason in context, e.g. he would not slow down by exactly the same amount on each strip; he may stop for breaks or to empty the grass box; he needs time to turn between strips; wet grass or slopes would change his speed
Do not accept vague answers such as "it is only a model" or "people are different".

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