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

Practise 17 exam-style questions on Arithmetic 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.

17 questions
3–9 marks each
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17 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 arithmetic series
\[50 + 47 + 44 + \dots\]
has sum \(S_n\) for its first \(n\) terms.

Find the greatest value of \(S_n\).

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

Model answer

The terms decrease by 3: \(u_n = 50 - 3(n - 1) = 53 - 3n\)
\(53 - 3n > 0\) when \(n < 17.7\), so \(u_{17} = 2\) is the last positive term and \(u_{18} = -1\)
\(S_n\) increases while the terms are positive, so the greatest value is \(S_{17}\):
\(S_{17} = \dfrac{17}{2}(50 + 2) = 442\)

Explanation

The sum keeps increasing while the terms being added are positive and starts decreasing once they are negative, so the greatest sum is the sum of all the positive terms. The 17th term, 2, is the last positive one, so the greatest sum is \(S_{17} = 442\). Treating \(S_n\) as a quadratic also works, but \(n\) must be a whole number.

Mark scheme

1 mark
M1 Finds the last positive term, \(50 - 3(n - 1) > 0\) giving \(n \leqslant 17\) (\(u_{17} = 2\), \(u_{18} = -1\)); or maximises \(S_n = \dfrac{n}{2}(103 - 3n)\) and chooses the nearest integer \(n = 17\); or a correct final answer of 442 with no working
An answer of 442 with no working scores 3 marks.
1 mark
M1 (dM1) Finds \(S_{17}\), e.g. \(\dfrac{17}{2}(50 + 2)\)
Depends on the previous M mark (with their integer value of \(n\)).
1 mark
A1 442
cao. Using the non-integer \(n = \dfrac{103}{6}\) gives 442.04: A0.

Question 2

5 marks

The arithmetic sequences \(P\) and \(Q\) are
\[P: \ 4, 11, 18, 25, \dots \qquad Q: \ 2, 7, 12, 17, 22, \dots\]
Show that the numbers that are terms of both \(P\) and \(Q\) form an arithmetic sequence, and state its first term and its common difference.

Find the sum of all the numbers less than 1000 that are terms of both \(P\) and \(Q\).

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

Model answer

\(P\): \(4 + 7(n - 1) = 7n - 3\); \(Q\): \(2 + 5(m - 1) = 5m - 3\)
A number is a term of both when \(7n - 3 = 5m - 3\), that is \(7n = 5m\). Since 7 and 5 have no common factor, \(n = 5k\) and \(m = 7k\) for a positive integer \(k\).
So the common terms are \(7(5k) - 3 = 35k - 3\): 32, 67, 102, ..., an arithmetic sequence with first term 32 and common difference 35.
\(35k - 3 < 1000\) gives \(k < 28.66\), so \(k \leqslant 28\): there are 28 such numbers, and the largest is \(35 \times 28 - 3 = 977\).
Sum \(= \dfrac{28}{2}(32 + 977) = 14 \times 1009 = 14126\)

Explanation

Equating the \(n\)th terms, \(7n - 3 = 5m - 3\), gives \(7n = 5m\), so the common terms are every 5th term of \(P\) (and every 7th term of \(Q\)); they start at 32 and go up by \(\mathrm{lcm}(7, 5) = 35\). There are 28 of them below 1000, the last being 977, so the sum is \(\dfrac{28}{2}(32 + 977) = 14126\). Common errors are taking 35 as the first common term and including 1012, which is over 1000.

Mark scheme

1 mark
M1 Writes the \(n\)th terms, \(7n - 3\) and \(5m - 3\), and equates them (\(7n = 5m\)); or lists terms of both sequences far enough to find two common terms, 32 and 67; or a correct final answer of 14 126 with no working (special case: 1 mark)
A correct answer of 14 126 with no working scores 1 mark (special case: the final A1); the common terms must be shown to form an arithmetic sequence for the first two marks, and the number of terms and the sum must be seen for the method marks.
1 mark
A1 The common terms are \(35k - 3\) for \(k = 1, 2, 3, \dots\): an arithmetic sequence with first term 32 and common difference 35, with a reason (e.g. \(7n = 5m\) needs \(n = 5k\) and \(m = 7k\); or the pattern repeats every \(\mathrm{lcm}(7, 5) = 35\))
Listing 32, 67, 102 with no reason why the pattern continues is A0.
1 mark
M1 Finds the number of common terms less than 1000: \(32 + 35(k - 1) < 1000\), so \(k \leqslant 28\) (the 28th is 977)
M1ft on their first term and common difference. Using \(\leqslant 1000\) gives the same answer here.
1 mark
M1 Uses the sum formula with their first term, common difference and number of terms: \(\dfrac{28}{2}(32 + 977)\) or \(\dfrac{28}{2}[2 \times 32 + 27 \times 35]\)
Not dependent on the previous M mark.
1 mark
A1 14 126
cao. Accept 14126. A common error is to include 1012 (29 terms), giving 15 138: A0.

Question 3

6 marks

Kai starts a savings plan. He saves £20 in month 1, £25 in month 2, £30 in month 3, and so on, saving £5 more each month than in the month before.

Find the month in which his total savings first exceed £3000, and the amount he saves in that month.

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

Model answer

\(S_n = \dfrac{n}{2}[2(20) + 5(n - 1)] = \dfrac{n}{2}(5n + 35)\)
\(\dfrac{n}{2}(5n + 35) > 3000\), so \(5n^2 + 35n - 6000 > 0\), i.e. \(n^2 + 7n - 1200 > 0\)
\(n^2 + 7n - 1200 = 0\) gives \(n = \dfrac{-7 + \sqrt{4849}}{2} = 31.3...\)
Check: \(S_{31} = 2945\) and \(S_{32} = 3120\), so the total first exceeds £3000 in month 32.
Amount saved in month 32: \(20 + 5(32 - 1) = 175\), i.e. £175

Explanation

The monthly amounts form an arithmetic sequence with \(a = 20\) and \(d = 5\), so set the sum of \(n\) terms greater than 3000 and solve the quadratic. Since \(n = 31.3\) is not a whole number of months, the total first passes £3000 in month 32, which can be checked with \(S_{31} = 2945\) and \(S_{32} = 3120\). The amount saved in month 32 is \(20 + 31 \times 5 = 175\); a common error is using \(32 \times 5\).

Mark scheme

1 mark
M1 Uses the sum formula with \(a = 20\) and \(d = 5\): \(\dfrac{n}{2}[40 + 5(n - 1)]\); or correct final answers of month 32 and £175 with no working (special case: 2 marks)
Answers of month 32 and £175 with no working score 2 marks (special case: the A1 for month 32 and the A1 for £175); the method marks need the sum and the \(n\)th term to be seen.
1 mark
A1 \(\dfrac{n}{2}[40 + 5(n - 1)] > 3000\), i.e. \(5n^2 + 35n - 6000 > 0\) oe
Accept an equation in place of the inequality.
1 mark
M1 Solves their three-term quadratic, e.g. \(n^2 + 7n - 1200 = 0\) gives \(n = 31.3...\), or evaluates the totals for 31 and 32 months (£2945 and £3120)
1 mark
A1 Month 32
An answer of 31 or 31.3 is A0.
1 mark
M1 Uses the \(n\)th term for their month: \(20 + 5(32 - 1)\)
A common error is \(20 + 5 \times 32 = 180\): M0.
1 mark
A1 £175
cao. Units are not required.

Question 4

8 marks

The seats in a theatre are arranged in rows. The front row has 20 seats, and each row after the first has 3 more seats than the row in front of it. The theatre has 1308 seats in total.

Find the number of rows in the theatre and the number of seats in the back row.

The theatre is to be extended by adding rows at the back, continuing the same pattern, so that it has at least 1800 seats. Find the smallest number of extra rows needed.

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

Model answer

\(S_n = \dfrac{n}{2}[2(20) + 3(n - 1)] = \dfrac{n}{2}(3n + 37)\)
\(\dfrac{n}{2}(3n + 37) = 1308\), so \(3n^2 + 37n - 2616 = 0\)
\((n - 24)(3n + 109) = 0\), so \(n = 24\) rows
Back row: \(20 + 3(24 - 1) = 89\) seats
At least 1800 seats: \(\dfrac{n}{2}(3n + 37) \geqslant 1800\)
\(3n^2 + 37n - 3600 = 0\) gives \(n = \dfrac{-37 + \sqrt{44569}}{6} = 29.01...\)
29 rows give \(\dfrac{29}{2}(87 + 37) = 1798\) seats and 30 rows give \(\dfrac{30}{2}(90 + 37) = 1905\) seats, so 30 rows are needed.
Extra rows: \(30 - 24 = 6\)

Explanation

The numbers of seats in the rows form an arithmetic sequence with \(a = 20\) and \(d = 3\), so the total for \(n\) rows is \(\dfrac{n}{2}(3n + 37)\); setting this equal to 1308 gives a quadratic with positive root 24. The back row is the 24th term, \(20 + 23 \times 3 = 89\). For at least 1800 seats the smallest whole number of rows is 30, because 29 rows give only 1798 seats, which means 6 extra rows; common errors are rounding 29.01 down and giving the total number of rows instead of the extra rows.

Mark scheme

1 mark
M1 Uses the sum formula with \(a = 20\) and \(d = 3\): \(\dfrac{n}{2}[40 + 3(n - 1)] = 1308\); or correct final answers of 24 rows, 89 seats and 6 extra rows with no working (special case: 3 marks)
Answers of 24 rows, 89 seats and 6 extra rows with no working score 3 marks (special case: the A1 for 24 rows, the B1 for 89 seats and the final A1 for 6 extra rows); the method marks need the equation and the inequality to be seen.
1 mark
A1 \(3n^2 + 37n - 2616 = 0\) oe
1 mark
M1 (dM1) Solves their three-term quadratic (factorising, the formula, completing the square or a calculator)
Depends on the first M mark. May be implied by \(n = 24\).
1 mark
A1 24 rows
\(n = -\dfrac{109}{3}\) must be rejected or not given.
1 mark
B1 89 seats in the back row (\(20 + 3 \times 23\))
B1ft on their number of rows: \(20 + 3(n - 1)\). A common error is \(20 + 3 \times 24 = 92\).
1 mark
M1 Solves \(\dfrac{n}{2}(3n + 37) \geqslant 1800\) (or \(= 1800\)), e.g. \(n = \dfrac{-37 + \sqrt{44569}}{6} = 29.01...\), or evaluates the totals for 29 and 30 rows
1 mark
A1 30 rows are needed (29 rows give 1798 seats; 30 rows give 1905)
An answer of 29 rows is A0: 29 rows give only 1798 seats.
1 mark
A1 6 extra rows
cao. Giving 30 as the final answer (the total number of rows, not the extra rows) is A0.

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