C2 June 2014 Q6
6. The first term of a geometric series is 20 and the common ratio is \(\dfrac{7}{8}\)
The sum to infinity of the series is \(S_\infty\)
The sum to \(N\) terms of the series is \(S_N\)
| Scheme | Marks |
|---|---|
| \(S_\infty = \dfrac{20}{1 - \frac{7}{8}}\ ;\ = 160\) | M1A1 |
| Accept correct answer only (160) | |
| (2) |
Notes
M1: Use of a correct \(S_\infty\) formula
A1: 160
| Scheme | Marks |
|---|---|
| \(S_{12} = \dfrac{20\left(1 - \left(\frac{7}{8}\right)^{12}\right)}{1 - \frac{7}{8}}\ ;\ = 127.77324\ldots\) | M1A1 |
| T & I in (b) requires all 12 terms to be calculated correctly for M1 and A1 for awrt 127.8 | |
| (2) |
Notes
M1: Use of a correct \(S_n\) formula with \(n = 12\) (condone missing brackets around 7/8)
A1: awrt 127.8
| Scheme | Marks |
|---|---|
| \(160 - \dfrac{20\left(1 - \left(\frac{7}{8}\right)^N\right)}{1 - \frac{7}{8}} \lt 0.5\) | M1 |
| \(160\left(\dfrac{7}{8}\right)^N \lt (0.5)\) or \(\left(\dfrac{7}{8}\right)^N \lt \left(\dfrac{0.5}{160}\right)\) | dM1 |
| \(N\log\left(\dfrac{7}{8}\right) \lt \log\left(\dfrac{0.5}{160}\right)\) | M1 |
| \(N \gt \dfrac{\log\left(\frac{0.5}{160}\right)}{\log\left(\frac{7}{8}\right)} = 43.19823\ldots \Rightarrow N = 44\) | A1 cso |
| (4) | |
| Total 8 |
Notes
1st M1: Applies \(S_N\) (GP only) with \(a = 20\), \(r = \tfrac{7}{8}\) and “uses” 0.5 and their \(S_\infty\) at any point in their working. (condone missing brackets around 7/8)(Allow \(=, \lt, \gt, \geqslant, \leqslant\)) but see note below.
dM1: Attempt to isolate \(+160\left(\tfrac{7}{8}\right)^N\) or \(+\left(\tfrac{7}{8}\right)^N\) oe (Allow \(=, \lt, \gt, \geqslant, \leqslant\)) but see note below. Dependent on the previous M1
M1: Uses the power law of logarithms or takes logs base 0.875 correctly to obtain an equation or an inequality of the form\[N\log\left(\frac{7}{8}\right) \lt \log\left(\frac{0.5}{\text{their } S_\infty}\right) \quad \text{or} \quad N \gt \log_{0.875}\left(\frac{0.5}{\text{their } S_\infty}\right)\](Allow \(=, \lt, \gt, \geqslant, \leqslant\)) but see note below.
A1 cso: \(N = 44\) (Allow \(N \geqslant 44\) but not \(N \gt 44\)
An incorrect inequality statement at any stage in a candidate’s working loses the final mark. Some candidates do not realise that the direction of the inequality is reversed in the final line of their solution. BUT it is possible to gain full marks for using =, as long as no incorrect working seen.
Trial & Improvement Method in (c):
1st M1: Attempts \(160 - S_N\) or \(S_N\) with at least one value for \(N \gt 40\)
2nd M1: Attempts \(160 - S_N\) or \(S_N\) with \(N = 43\) or \(N = 44\)
3rd M1: For evidence of examining \(160 - S_N\) or \(S_N\) for both \(N = 43\) and \(N = 44\) with both values correct to 2 DP
Eg: \(160 - S_{43} =\) awrt 0.51 and \(160 - S_{44} =\) awrt 0.45
or \(S_{43} =\) awrt 159.49 and \(S_{44} =\) awrt 159.55
A1: \(N = 44\) cso
Answer of \(N = 44\) only with no working scores no marks