June 2024 Paper 1 Q11
11 The first three terms of a geometric sequence are \(5k - 2\), \(3k - 6\), \(k + 2\), where \(k\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| If geometric, then \(r = \frac{3k-6}{5k-2}\) | M1 | 2.4 |
| Common ratio gives \(\dfrac{3k-6}{5k-2} = \dfrac{k+2}{3k-6}\) So \(9k^2 - 36k + 36 = 5k^2 + 8k - 4\) | M1 | 2.1 |
| So \(k^2 - 11k + 10 = 0\) | A1 | 2.1 |
| [3] |
Notes
M1: Allow instead for \(r = \frac{k+2}{3k-6}\) or \(r^2 = \frac{k+2}{5k-2}\) in any form Soi
M1: Forms an equation in \(k\) which need not be simplified
A1: AG Rearranges to correct three term quadratic.
At least one intermediate step must be shown.
SC1: for showing \(k = 1\) leads to \(3, -3, 3\) \((r = -1)\) and that \(k = 10\) leads to 48, 24, 12 \(\left(r = \frac{1}{2}\right)\) and demonstrating that both are geometric
| Scheme | Marks | AO |
|---|---|---|
| So \(k = 1, 10\) When \(k = 1\) the sum of 20 terms is \((3 + (-3)) + (3 + (-3)) + \ldots + (3 + (-3)) = 0\) | M1 M1 A1 | 1.1a 3.1a 1.1 |
| [3] |
Notes
M1: Solves the quadratic to give at least one root Soi
M1: Evaluating the terms of the sequence when \(k = 1\)
A1: cao
Alternative for the last 2 marks
| Scheme | Marks | AO |
|---|---|---|
| \(S_{20} = \dfrac{3(1 - (-1)^{20})}{1 - (-1)} = 0\) | M1 A1 |
M1: Using the formula for the sum of terms of a GP with \(r = -1\)
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| When \(k = 10\) the sequence is 48, 24, 12… So \(a = 48, r = \dfrac{1}{2}\) | B1 | 3.1a |
| \(S_\infty = \dfrac{48}{1 - \frac{1}{2}} = 96\) | B1 | 1.1 |
| [2] |
Notes
B1: Identifies the first term and common ratio soi
B1: cao