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P4.2Sequences and recurrence relations

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

Practise Sequences and recurrence relations. 1 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 1Easy5 marks
A sequence \(u_1, u_2, u_3, \ldots\) is defined by
\(u_{1} = 1, \qquad u_{n+1} = 2u_n - 3, \quad n \ge 1\).
(a) Find the values of \(u_{2}\), \(u_{3}\) and \(u_{4}\).[2]
(b) Find \(\sum_{r=1}^{4} u_r\).[2]
(c) State whether the sequence is increasing, decreasing or neither.[1]
Show the answer and mark scheme
(a) Answer: \(u_{2} = -1, \ u_{3} = -5, \ u_{4} = -13\)
  • M1 for a correct use of the recurrence relation at least once, e.g. \(u_{2} = 2 \times 1 - 3\)
  • A1 for all three values correct

Worked solution: \(u_{2} = 2 \times 1 - 3 = -1\), \(u_{3} = 2 \times (-1) - 3 = -5\), \(u_{4} = 2 \times (-5) - 3 = -13\).

(b) Answer: \(-18\)
  • M1 for adding their first four terms, \(u_{1} + u_{2} + u_{3} + u_{4}\)
  • A1ft for \(-18\), following through their terms

Worked solution: \(\sum_{r=1}^{4} u_r = 1 + (-1) + (-5) + (-13) = -18\)

(c) Answer: Decreasing
  • B1 for decreasing

Worked solution: Subtracting consecutive relations gives \(u_{n+2} - u_{n+1} = 2(u_{n+1} - u_{n})\), so the differences between consecutive terms are \(-2, -4, -8, \ldots\).
Every difference is negative, so the sequence is decreasing.

Question 2Medium6 marks
A sequence \(a_1, a_2, a_3, \ldots\) is defined by
\(a_{1} = -1, \qquad a_{n+1} = -2a_n + c, \quad n \ge 1\)
where \(c\) is a constant.
(a) Find an expression for \(a_{2}\) in terms of \(c\).[1]
(b) Show that \(a_{3} = -c - 4\).[2]
(c) Given that \(\sum_{r=1}^{4} a_r = -10\), find the value of \(c\).[3]
Show the answer and mark scheme
(a) Answer: \(a_{2} = 2 + c\)
  • B1 for \(2 + c\) oe

Worked solution: \(a_{2} = -2 \times (-1) + c = 2 + c\)

(b) Answer: Shown
  • M1 for \(a_{3} = -2(2 + c) + c\)
  • A1* for \(-c - 4\) with no errors (cso)

Worked solution: \(a_{3} = -2(2 + c) + c = -2c - 4 + c = -c - 4\)

(c) Answer: \(c = -5\)
  • M1 for \(a_{4} = -2(-c - 4) + c\) (\(= 8 + 3c\))
  • M1 for adding their four terms and setting the sum equal to \(-10\), then solving
  • A1 for \(c = -5\)

Worked solution: \(a_{4} = -2(-c - 4) + c = 8 + 3c\)
\(\sum_{r=1}^{4} a_r = -1 + (2 + c) + (-c - 4) + (8 + 3c) = 5 + 3c\)
\(5 + 3c = -10\), so \(3c = -15\), giving \(c = -5\).

Question 3Hard9 marks
A sequence is defined by
\[u_{n + 1} = \frac{1}{1 - u_n}, \quad n \ge 1\]
(a) Given that \(u_1 = 2\), find \(u_2\), \(u_3\) and \(u_4\).[2]
(b) Prove that, for any value of \(u_1\) other than 0 and 1, \(u_4 = u_1\).[4]
(c) Explain why the sequence is not defined when \(u_1 = 0\).[1]
(d) Given that \(u_1 = 2\), find \(\displaystyle\sum_{r=1}^{100} u_r\).[2]
Show the answer and mark scheme
(a) Answer: \(u_2 = -1\), \(u_3 = \frac{1}{2}\), \(u_4 = 2\)
  • M1 for using the recurrence at least once correctly
  • A1 for \(-1, \frac{1}{2}, 2\)

Worked solution: \(u_2 = \frac{1}{1 - 2} = -1\), \(u_3 = \frac{1}{1 - (-1)} = \frac{1}{2}\), \(u_4 = \frac{1}{1 - \frac{1}{2}} = 2\).

(b) Answer: \(u_2 = \frac{1}{1 - u_1}\), \(u_3 = \frac{u_1 - 1}{u_1}\), \(u_4 = \frac{1}{1 - \frac{u_1 - 1}{u_1}} = \frac{u_1}{1} = u_1\).
  • M1 for \(u_2 = \frac{1}{1 - u_1}\) and an attempt at \(u_3\)
  • A1 for \(u_3 = \frac{1 - u_1}{-u_1}\) or \(\frac{u_1 - 1}{u_1}\)
  • M1 for \(u_4 = \dfrac{1}{1 - \frac{u_1 - 1}{u_1}}\) and simplifying
  • A1* for \(u_4 = u_1\)

Worked solution: Let \(u_1 = a\). \(u_2 = \dfrac{1}{1 - a}\).
\(u_3 = \dfrac{1}{1 - \frac{1}{1 - a}} = \dfrac{1 - a}{(1 - a) - 1} = \dfrac{1 - a}{-a} = \dfrac{a - 1}{a}\).
\(u_4 = \dfrac{1}{1 - \frac{a - 1}{a}} = \dfrac{a}{a - (a - 1)} = a\). So \(u_4 = u_1\): the sequence is periodic with period 3.

(c) Answer: \(u_2 = 1\), and then \(u_3 = \frac{1}{0}\) is not defined.
  • B1 for \(u_2 = 1\), so \(u_3\) would need division by zero

Worked solution: If \(u_1 = 0\), then \(u_2 = \frac{1}{1 - 0} = 1\) and \(u_3 = \frac{1}{1 - 1}\), which is not defined.

(d) Answer: 51.5
  • M1 for using the period: \(33\) complete cycles with sum \(2 - 1 + \frac{1}{2} = \frac{3}{2}\) each, plus \(u_{100} = u_1 = 2\)
  • A1 for 51.5

Worked solution: The terms repeat \(2, -1, \frac{1}{2}\) with cycle sum \(\frac{3}{2}\). \(100 = 3 \times 33 + 1\), so the sum is \(33 \times \frac{3}{2} + u_{100} = 49.5 + 2 = 51.5\).

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