| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing standard geometric and arithmetic series formulas with routine calculations. Part (a) involves verifying a given common ratio (division), applying the sum to infinity formula, and writing the nth term—all direct applications of memorized formulas. Part (b) similarly requires basic arithmetic series formulas. No problem-solving insight or multi-step reasoning is needed, making this easier than average but not trivial since it requires accurate formula recall and arithmetic across multiple parts. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{294}{420} = 0.7\) | B1 (1 mark) | AG. Accept any valid justification to the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{a}{1-r} = \frac{420}{1-0.7}\) | M1 | \(\frac{a}{1-r}\) used |
| \(S_\infty = 1400\) | A1 (2 marks) | 1400 NMS mark as 2/2 or 0/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(n\)th term \(= 600 \times (0.7)^n\) | B2 (2 marks) | If not B2 award B1 for \(420 \times (0.7)^{n-1}\) OE |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{u_n\} = 248 - 8n\) | B1 (1 mark) | Accept ACF |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_k = 0 \Rightarrow 8k = 248\) | M1 | \(248 - 8k = 0\) OE e.g. \(240 + (k-1)(-8) = 0\); ft if no recovery, on c's (b)(i) answer |
| \(k = 31\) | A1 | |
| \(\sum_{n=1}^{k} u_n = 240 + 232 + \ldots + 0 = \frac{k}{2}[240 + 0]\) | M1 | For \(\frac{k}{2}[240 + 0]\) or for \(\frac{k}{2}[\text{c's } u_1 + 0]\); OE e.g. \(\frac{k}{2}[2 \times \text{c's } u_1 + (k-1)(-8)]\) |
| \(\sum_{n=1}^{k} u_n = 3720\) | A1 (4 marks) | 3720 |
# Question 6:
## Part 6(a)(i):
$r = \frac{294}{420} = 0.7$ | B1 (1 mark) | AG. Accept any valid justification to the given answer
## Part 6(a)(ii):
$S_\infty = \frac{a}{1-r} = \frac{420}{1-0.7}$ | M1 | $\frac{a}{1-r}$ used
$S_\infty = 1400$ | A1 (2 marks) | 1400 NMS mark as 2/2 or 0/2
## Part 6(a)(iii):
$n$th term $= 600 \times (0.7)^n$ | B2 (2 marks) | If not B2 award B1 for $420 \times (0.7)^{n-1}$ OE
## Part 6(b)(i):
$\{u_n\} = 248 - 8n$ | B1 (1 mark) | Accept ACF
## Part 6(b)(ii):
$u_k = 0 \Rightarrow 8k = 248$ | M1 | $248 - 8k = 0$ OE e.g. $240 + (k-1)(-8) = 0$; ft if no recovery, on c's (b)(i) answer
$k = 31$ | A1 |
$\sum_{n=1}^{k} u_n = 240 + 232 + \ldots + 0 = \frac{k}{2}[240 + 0]$ | M1 | For $\frac{k}{2}[240 + 0]$ or for $\frac{k}{2}[\text{c's } u_1 + 0]$; OE e.g. $\frac{k}{2}[2 \times \text{c's } u_1 + (k-1)(-8)]$
$\sum_{n=1}^{k} u_n = 3720$ | A1 (4 marks) | 3720
---6
\begin{enumerate}[label=(\alph*)]
\item A geometric series begins $420 + 294 + 205.8 + \ldots$.
\begin{enumerate}[label=(\roman*)]
\item Show that the common ratio of the series is 0.7 .
\item Find the sum to infinity of the series.
\item Write the $n$th term of the series in the form $p \times q ^ { n }$, where $p$ and $q$ are constants.
\end{enumerate}\item The first term of an arithmetic series is 240 and the common difference of the series is - 8 .
The $n$th term of the series is $u _ { n }$.
\begin{enumerate}[label=(\roman*)]
\item Write down an expression for $u _ { n }$.
\item Given that $u _ { k } = 0$, find the value of $\sum _ { n = 1 } ^ { k } u _ { n }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q6 [10]}}