June 2024 Paper 1 Q10
10 In this question you must show detailed reasoning.
The first three terms of a convergent geometric progression are \(2x + 3\), \(x + 9\) and \(2x - 6\) respectively.
Determine the sum to infinity of this geometric progression. [8]
| Scheme | Marks | AO |
|---|---|---|
| DR \(r(2x + 3) = x + 9;\ \ r(x + 9) = 2x - 6;\ \ r^2(2x + 3) = 2x - 6\) | B1 | 3.1a |
| \(\dfrac{x + 9}{2x + 3} = \dfrac{2x - 6}{x + 9}\) oe | M1* | 3.1a |
| \(x^2 + 18x + 81 = 4x^2 - 12x + 6x - 18\) \(3x^2 - 24x - 99 = 0\) | A1 | 1.1 |
| \((x - 11)(x + 3) = 0\) \(x = 11,\ x = -3\) | A1 | 2.1 |
| \(r = \tfrac{4}{5},\ r = -2\) | A1 | 1.1 |
| \(S_\infty = \dfrac{a}{1 - r} = \dfrac{25}{1 - \frac{4}{5}}\) | M1d* | 3.2a |
| \(S_\infty = 125\) | A1 | 1.1 |
| \(S_\infty\) only exists for \(|r| \lt 1\), so \(r = -2\) is not a valid solution | B1 | 2.5 |
| [8] |
Notes
B1: Obtain any correct equation in terms of \(r\) and \(x\)
Could be implied by later work
May use other than \(r\)
M1*: Attempt equation in terms of only \(x\)
This equation would imply the B1
Or correct equation in terms of only \(r\)
A1: Obtain any correct equation not involving fractions or brackets
May still have like terms not yet combined
May result in a cubic depending on method (probably \(6x^3 - 39x^2 - 270x - 297 = 0\))
A1: Solve quadratic BC to obtain both correct \(x\) values
Or solve cubic, to obtain three correct roots (third is likely to be \(x = -1.5\))
A1: Obtain at least \(r = \tfrac{4}{5}\)
If second value of \(r\) given then it must be correct (if third value given then it must be consistent with their correct cubic roots)
M1d*: Attempt sum to infinity, using correct formula, with their \(r\) and their attempt at \(a\)
Must be using their numerical values for \(a\) and \(r\) with \(|r| \lt 1\)
ISW using additional value(s) of \(r\)
M0 if using their \(x\) and not attempt at \(a\)
A1: Obtain 125 only
A0 if additional solution
B1: Clear explanation as to why \(r = -2\) is discarded
Must be considering correct \(r\) value, so B0 if rejecting \(x = -3\) as \(|-3| \gt 1\)
Could generate the terms \(-3, 6, (-12)\) and hence conclude with ‘divergent sequence’
If additional solutions for \(x\) and/or \(r\) from cubic then they must also be correct and explicitly rejected
NB Eliminating \(x\) not \(r\) is a valid method, and could gain full credit. When solving their quadratic there is no need to see \(r = -2\) (and hence \(x = -3\)) to award the A marks