| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Easy -1.2 This is a straightforward application of standard geometric series formulas with no problem-solving required. All three parts involve direct substitution into well-known formulas (nth term, sum to infinity, and finite sum) with simple arithmetic using r=1/2. This is easier than average as it's purely procedural recall. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks |
|---|---|
| \(20\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{S_\infty\} = \frac{a}{1-r} = \frac{80}{1-\frac{1}{2}}\) | M1 | \(\frac{a}{1-r}\) used with \(a = 80\) and \(r = 0.5\) OE |
| \(\{S_\infty\} = 160\) | A1 | NMS 160 gets 2 marks unless rounding seen |
| \(\{S_{12}\} = \frac{80(1-r^{12})}{1-r} = 160(1-0.5^{12})\) | M1 | \(\frac{80(1-r^{12})}{1-r}\) seen (or used with \(r=0.5\) OE) |
| \(= 159.96(0937.) = 159.96\) to 2dp | A1 | Condone > 2dp |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 5 |
**1(a)**
$20$ | B1 |
---|---|---
**1(b)**
$\{S_\infty\} = \frac{a}{1-r} = \frac{80}{1-\frac{1}{2}}$ | M1 | $\frac{a}{1-r}$ used with $a = 80$ and $r = 0.5$ OE
$\{S_\infty\} = 160$ | A1 | NMS 160 gets 2 marks unless rounding seen
| |
$\{S_{12}\} = \frac{80(1-r^{12})}{1-r} = 160(1-0.5^{12})$ | M1 | $\frac{80(1-r^{12})}{1-r}$ seen (or used with $r=0.5$ OE)
$= 159.96(0937.) = 159.96$ to 2dp | A1 | Condone > 2dp
**1(c)**
Total | | 5
---1 A geometric series has first term 80 and common ratio $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the third term of the series.
\item Find the sum to infinity of the series.
\item Find the sum of the first 12 terms of the series, giving your answer to two decimal places.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q1 [5]}}