C2 June 2006 Q9
9. A geometric series has first term \(a\) and common ratio \(r\).
The second term of the series is 4 and the sum to infinity of the series is 25.
Given that \(r\) takes the larger of its two possible values,
| Scheme | Marks |
|---|---|
| \(ar = 4, \quad \dfrac{a}{1 - r} = 25\) (These can be seen elsewhere) | B1, B1 |
| \(a = 25(1 - r) \qquad 25r(1 - r) = 4\) M: Eliminate \(a\) | M1 |
| \(25r^2 - 25r + 4 = 0\) (*) | A1cso |
| (4) |
Notes
The M mark is not dependent, but both expressions must contain both \(a\) and \(r\).
| Scheme | Marks |
|---|---|
| \((5r - 1)(5r - 4) = 0 \quad r = \ldots, \qquad \dfrac{1}{5}\) or \(\dfrac{4}{5}\) | M1, A1 |
| (2) |
Notes
Special case:
One correct \(r\) value given, with no method (or perhaps trial and error): B1 B0.
| Scheme | Marks |
|---|---|
| \(r = \ldots \Rightarrow a = \ldots, \qquad 20\) or 5 | M1, A1 |
| (2) |
Notes
M1: Substitute one \(r\) value back to find a value of \(a\).
| Scheme | Marks |
|---|---|
| \(S_n = \dfrac{a(1 - r^n)}{1 - r}\), but \(\dfrac{a}{1 - r} = 25\), so \(S_n = 25(1 - r^n)\) (*) | B1 |
| (1) |
Notes
Sufficient here to verify with just one pair of values of \(a\) and \(r\).
| Scheme | Marks |
|---|---|
| \(25(1 - 0.8^n) \gt 24\) and proceed to \(n = \ldots\) (or \(\gt\), or \(\lt\)) with no unsound algebra. | M1 |
| \(\left(n \gt \dfrac{\log 0.04}{\log 0.8} \quad (= 14.425\ldots)\right) \qquad n = 15\) | A1 |
| (2) | |
| (11 marks) |
Notes
Accept “=” rather than inequalities throughout, and also allow the wrong inequality to be used at any stage.
M1 requires use of their larger value of \(r\).
A correct answer with no working scores both marks.
For “trial and error” methods, to score M1, a value of \(n\) between 12 and 18 (inclusive) must be tried.