| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| 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 geometric series question requiring only direct substitution and standard formula application. Parts (a)-(c) are routine calculations, while part (d) adds a minor twist of finding a partial sum to infinity, but this is still a standard technique (find S_∞ minus S_3). No problem-solving or novel insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_2 = 48\times\frac{1}{16} = 3\) | B1, B1F | CAO must be 12; if not correct, ft on c's \(u_1\times\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{1}{4}\) | B1F | Only ft on \(r = (\text{c's } u_2)\div(\text{c's } u_1)\) if \( |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{u_1}{1-r} = \frac{12}{1-\frac{1}{4}} = 16\) | M1, A1F | Use of \(\frac{a}{1-r}\); ft on c's \(u_1\) and \(r\) provided \( |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{n=4}^{\infty}u_n = 0.25\) | M1, B1, A1 | OE e.g. RHS \(S_\infty-(u_1+u_2+u_3)\); either result, or better e.g. \(\sum_{n=1}^3 u_n=15.75\); NMS scores 0/3 |
| Alternative: \(\sum_{n=4}^{\infty}u_n = \frac{u_4}{1-r}\), \(u_4=\frac{3}{16}\), \(\sum_{n=4}^{\infty}u_n = \frac{3}{16}\div\frac{3}{4}=\frac{1}{4}\) | (M1)(B1)(A1) | NMS scores 0/3 |
# Question 4:
## Part (a)
$u_1 = 12$
$u_2 = 48\times\frac{1}{16} = 3$ | B1, B1F | CAO must be 12; if not correct, ft on c's $u_1\times\frac{1}{4}$
## Part (b)
$r = \frac{1}{4}$ | B1F | Only ft on $r = (\text{c's } u_2)\div(\text{c's } u_1)$ if $|r|<1$; answers may be in equivalent fraction or exact decimal form
## Part (c)
$S_\infty = \frac{u_1}{1-r} = \frac{12}{1-\frac{1}{4}} = 16$ | M1, A1F | Use of $\frac{a}{1-r}$; ft on c's $u_1$ and $r$ provided $|r|<1$; if not 16, ft on c's $u_1$ and $r$ provided $|r|<1$
## Part (d)
$\sum_{n=4}^{\infty}u_n = S_\infty - \sum_{n=1}^{3}u_n$
$u_3 = \frac{3}{4}$ (or $\sum_{n=1}^3 u_n = \frac{12(1-0.25^3)}{1-0.25}$)
$\sum_{n=4}^{\infty}u_n = 0.25$ | M1, B1, A1 | OE e.g. RHS $S_\infty-(u_1+u_2+u_3)$; either result, or better e.g. $\sum_{n=1}^3 u_n=15.75$; NMS scores 0/3
Alternative: $\sum_{n=4}^{\infty}u_n = \frac{u_4}{1-r}$, $u_4=\frac{3}{16}$, $\sum_{n=4}^{\infty}u_n = \frac{3}{16}\div\frac{3}{4}=\frac{1}{4}$ | (M1)(B1)(A1) | NMS scores 0/34 The $n$th term of a geometric series is $u _ { n }$, where $u _ { n } = 48 \left( \frac { 1 } { 4 } \right) ^ { n }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $u _ { 1 }$ and the value of $u _ { 2 }$.
\item Find the value of the common ratio of the series.
\item Find the sum to infinity of the series.
\item Find the value of $\sum _ { n = 4 } ^ { \infty } u _ { n }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q4 [8]}}