C1 June 2013 Q4
4. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned} a_1 &= 4 \\ a_{n+1} &= k(a_n + 2), \qquad \text{for } n \geqslant 1 \end{aligned}\]where \(k\) is a constant.
Given that \(\displaystyle\sum_{i=1}^{3} a_i = 2\),
For this question, mark (a) and (b) together and ignore labelling.
| Scheme | Marks |
|---|---|
| \((a_2 =)\ k(4 + 2) \quad (= 6k)\) | B1 |
| (1) |
Notes
B1: Any correct (possibly un-simplified) expression
| Scheme | Marks |
|---|---|
| \(a_3 = k(\text{their } a_2 + 2) \quad (= 6k^2 + 2k)\) | M1 |
| \(a_1 + a_2 + a_3 = 4 + (6k) + (6k^2 + 2k)\) | M1 |
| \(4 + (6k) + (6k^2 + 2k) = 2\) | A1 |
| Solves \(6k^2 + 8k + 2 = 0\) to obtain \(k =\) \((6k^2 + 8k + 2 = 2(3k + 1)(k + 1))\) | M1 |
| \(k = -1/3\) | A1 |
| \(k = -1\) | B1 |
| (6) | |
| (7 marks) |
Notes
M1: An attempt at \(a_3\). Can follow through their answer to (a) but \(a_2\) must be an expression in \(k\).
M1: An attempt to find their \(a_1 + a_2 + a_3\)
A1: A correct equation in any form.
M1: Solves their 3TQ as far as \(k =\) .... according to the general principles. (An independent mark for solving their three term quadratic)
A1: Any equivalent fraction
B1: Must be from a correct equation. (Do not accept un-simplified)
Note that it is quite common to think the sequence is an AP. Unless they find \(a_3\), this is likely only to score the M1 for solving their quadratic.