| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find N for S_∞ - S_N condition |
| Difficulty | Standard +0.3 This is a straightforward geometric series question requiring standard formula application. Part (a) uses the sum to infinity formula, part (b) applies the finite sum formula, and part (c) requires solving an inequality—all routine C2 techniques with no novel insight needed. Slightly above average difficulty only due to the algebraic manipulation in part (c). |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_\infty = \frac{20}{1-\frac{7}{8}} = 160\) | M1 | Use of correct \(S_\infty\) formula |
| \(= 160\) | A1 | Accept correct answer only (160) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{12} = \frac{20(1-(\frac{7}{8})^{12})}{1-\frac{7}{8}} = 127.77324...\) | M1 | Use of correct \(S_n\) formula with \(n=12\); condone missing brackets around 7/8 |
| \(= 127.8\) | A1 | awrt 127.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(160 - \frac{20(1-(\frac{7}{8})^N)}{1-\frac{7}{8}} < 0.5\) | M1 | Applies \(S_N\) (GP only) with \(a=20\), \(r=\frac{7}{8}\) and uses 0.5 and their \(S_\infty\); condone missing brackets; allow \(=,<,>,\geq,\leq\) |
| \(160\left(\frac{7}{8}\right)^N < (0.5)\) or \(\left(\frac{7}{8}\right)^N < \left(\frac{0.5}{160}\right)\) | dM1 | Attempt to isolate \(+160\left(\frac{7}{8}\right)^N\) or \(+\left(\frac{7}{8}\right)^N\); dependent on previous M1 |
| \(N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{160}\right)\) | M1 | Uses power law of logarithms correctly to obtain equation or inequality of the form \(N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{\text{their }S_\infty}\right)\) or \(N > \log_{0.875}\left(\frac{0.5}{\text{their }S_\infty}\right)\) |
| \(N > \frac{\log(\frac{0.5}{160})}{\log(\frac{7}{8})} = 43.19823... \Rightarrow N = 44\) | A1cso | \(N=44\); allow \(N \geq 44\) but not \(N > 44\) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_\infty = \frac{20}{1-\frac{7}{8}} = 160$ | M1 | Use of correct $S_\infty$ formula |
| $= 160$ | A1 | Accept correct answer only (160) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{12} = \frac{20(1-(\frac{7}{8})^{12})}{1-\frac{7}{8}} = 127.77324...$ | M1 | Use of correct $S_n$ formula with $n=12$; condone missing brackets around 7/8 |
| $= 127.8$ | A1 | awrt 127.8 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $160 - \frac{20(1-(\frac{7}{8})^N)}{1-\frac{7}{8}} < 0.5$ | M1 | Applies $S_N$ (GP only) with $a=20$, $r=\frac{7}{8}$ and uses 0.5 and their $S_\infty$; condone missing brackets; allow $=,<,>,\geq,\leq$ |
| $160\left(\frac{7}{8}\right)^N < (0.5)$ or $\left(\frac{7}{8}\right)^N < \left(\frac{0.5}{160}\right)$ | dM1 | Attempt to isolate $+160\left(\frac{7}{8}\right)^N$ or $+\left(\frac{7}{8}\right)^N$; dependent on previous M1 |
| $N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{160}\right)$ | M1 | Uses power law of logarithms correctly to obtain equation or inequality of the form $N\log\left(\frac{7}{8}\right) < \log\left(\frac{0.5}{\text{their }S_\infty}\right)$ or $N > \log_{0.875}\left(\frac{0.5}{\text{their }S_\infty}\right)$ |
| $N > \frac{\log(\frac{0.5}{160})}{\log(\frac{7}{8})} = 43.19823... \Rightarrow N = 44$ | A1cso | $N=44$; allow $N \geq 44$ but not $N > 44$ |
---6. The first term of a geometric series is 20 and the common ratio is $\frac { 7 } { 8 }$
The sum to infinity of the series is $S _ { \infty }$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $S _ { \infty }$
The sum to $N$ terms of the series is $S _ { N }$
\item Find, to 1 decimal place, the value of $S _ { 12 }$
\item Find the smallest value of $N$, for which
$$S _ { \infty } - S _ { N } < 0.5$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2014 Q6 [8]}}