June 2023 Paper 1 Q11
11 The \(n\)th term of a sequence is \(u_n\)
The sequence is defined by
\[u_{n+1} = pu_n + 70\]where \(u_1 = 400\) and \(p\) is a constant.
(a) Find an expression, in terms of \(p\), for \(u_2\) [1 mark]
(b) It is given that \(u_3 = 382\)
(i) Show that \(p\) satisfies the equation\[200p^2 + 35p - 156 = 0\] [3 marks]
(ii) It is given that the sequence is a decreasing sequence.
Find the value of \(u_4\) and the value of \(u_5\) [3 marks]
(c) The limit of \(u_n\) as \(n\) tends to infinity is \(L\)
(i) Write down an equation for \(L\) [1 mark]
(ii) Find the value of \(L\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(400p + 70\) | B1 | 1.1b |
| (1) |
Typical solution
\[400p + 70\]| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes 382 or their \(u_2\) into \(u_3 = pu_2 + 70\) | M1 | 1.1a |
| Substitutes 382 and their \(u_2\) into \(u_3 = pu_2 + 70\) To obtain a quadratic equation in p PI by \(382 = p(400p + 70) + 70\) | M1 | 3.1a |
| Obtains correct equation and rearranges to obtain given answer. Must see brackets expanded before given answer. | R1 | 2.1 |
| (3) | ||
| (ii) Obtains both \(p\) = 0.8 and -0.975 PI by correct \(u_4 = 375.6\) and \(u_5 = 370.48\) | B1 | 1.1b |
| Uses \(p\) = 0.8 or -0.975 to obtain a value for \(u_4\) PI by \(375.6\), \(-302.45\), \(370.48\) Accept equivalent fractions or AWRT \(364.89\) | M1 | 3.1a |
| Deduces correct values for \(u_4\) and \(u_5\). \((u_4 =)\,375.6\) and \((u_5 =)\,370.48\) Accept equivalent fractions If incorrect values are seen they must be rejected. | R1 | 2.2a |
| (3) |
Typical solution
(i)
\[382 = pu_2 + 70\]\[382 = p(400p + 70) + 70\]\[382 = 400p^2 + 70p + 70\]\[400p^2 + 70p - 312 = 0\]\[200p^2 + 35p - 156 = 0\](ii)
\[p = 0.8,\quad p = -0.975\]\[p = -0.975\]\[\Rightarrow u_4 = -302.45,\ u_5 = 364.88875\]not decreasing
\[p = 0.8\]\[\Rightarrow u_4 = 375.6,\ u_5 = 370.48\]| Scheme | Marks | AO |
|---|---|---|
| (i) Forms the equation \(L = pL + 70\) or \((1 - p)L = 70\) Or with \(p\) = 0.8 or -0.975 substituted into either of these equations accept if \(1 - p\) is evaluated ISW | B1 | 3.1a |
| (1) | ||
| (ii) Deduces the value of \(L\) is 350 or AWRT 35.4 Accept \(\dfrac{2800}{79}\) or both | R1 | 2.2a |
| (1) | ||
| (9 marks) |
Typical solution
(i)
\[L = 0.8L + 70\](ii)
\[350\]