| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find first term from conditions |
| Difficulty | Standard +0.3 This is a straightforward multi-part geometric series question requiring standard formulas (sum to infinity, nth term) and basic logarithm laws. Part (a) is simple algebra, (b) uses the standard sum formula, and (c) involves routine manipulation of logarithms with clear scaffolding. All techniques are standard C2 content with no novel insight required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | \(\frac{a}{1-r} = 4a\) ⟹ \(1 - r = \frac{a}{4a}\) or a = 4a(1−r); \(1 - r = \frac{1}{4}\) ⟹ \(r = \frac{3}{4}\) | M1, A1, A1 |
| 5(b) | \((S_{10}) = \frac{48(1-0.75^{10})}{1-r}\) = 192(1−0.75¹⁰) = 181.1878 to 4dp | M1, A1 |
| 5(c)(i) | \(u_n = ar^{n-1} = a(\frac{3}{4})^{n-1} = 48(\frac{3}{4})^{n-1}\); \(u_{2n} = ar^{2n-1} = a(\frac{3}{4})^{2n-1} = 48(\frac{3}{4})^{2n-1}\) | B1, B1 ft |
| 5(c)(ii) | \(\frac{u_n}{u_{2n}} = \frac{ar^{n-1}}{ar^{2n-1}} = \frac{r^{n-1}}{r^{2n-1}}\); \(\log_{10} u_n - \log_{10} u_{2n} = \log_{10}\frac{u_n}{u_{2n}}\) = \(\log_{10}(r^{-n}) = -n\log_{10}\frac{3}{4} = n\log_a\frac{4}{3}\) | M1, M1, A1 |
| 5(c)(iii) | \(\log_{10}[\frac{u_{100}}{u_{200}}] = 100\log_{10}(\frac{4}{3})\) = 12.49...., = 12.5 to 3 sf | M1, A1 |
5(a) | $\frac{a}{1-r} = 4a$ ⟹ $1 - r = \frac{a}{4a}$ or a = 4a(1−r); $1 - r = \frac{1}{4}$ ⟹ $r = \frac{3}{4}$ | M1, A1, A1 | (Accept $S_{\infty} = \frac{a}{1-\frac{3}{4}}$); Either (or better) (or $S_{\infty} = 4a$ if M1 above); AG CSO Be convinced. (or statement 4 times 1st term) |
5(b) | $(S_{10}) = \frac{48(1-0.75^{10})}{1-r}$ = 192(1−0.75¹⁰) = 181.1878 to 4dp | M1, A1 | Correct formula with n = 10 and one of a = 48 or $r = \frac{3}{4}$ OE |
5(c)(i) | $u_n = ar^{n-1} = a(\frac{3}{4})^{n-1} = 48(\frac{3}{4})^{n-1}$; $u_{2n} = ar^{2n-1} = a(\frac{3}{4})^{2n-1} = 48(\frac{3}{4})^{2n-1}$ | B1, B1 ft | ft on candidate's $u_n = ar^{\text{function of n}}$ |
5(c)(ii) | $\frac{u_n}{u_{2n}} = \frac{ar^{n-1}}{ar^{2n-1}} = \frac{r^{n-1}}{r^{2n-1}}$; $\log_{10} u_n - \log_{10} u_{2n} = \log_{10}\frac{u_n}{u_{2n}}$ = $\log_{10}(r^{-n}) = -n\log_{10}\frac{3}{4} = n\log_a\frac{4}{3}$ | M1, M1, A1 | Eliminating a (or 48) or log a; Using at least one log law; AG CSO Full valid completion |
5(c)(iii) | $\log_{10}[\frac{u_{100}}{u_{200}}] = 100\log_{10}(\frac{4}{3})$ = 12.49...., = 12.5 to 3 sf | M1, A1 | AG CSO Be convinced; SC:Those applying 'hence' to (i) rather than to (ii) Mark as B2 |5 The sum to infinity of a geometric series is four times the first term of the series.
\begin{enumerate}[label=(\alph*)]
\item Show that the common ratio, $r$, of the geometric series is $\frac { 3 } { 4 }$.
\item The first term of the geometric series is 48 . Find the sum of the first 10 terms of the series, giving your answer to four decimal places.
\item The $n$th term of the geometric series is $u _ { n }$ and the ( $2 n$ )th term of the series is $u _ { 2 n }$.
\begin{enumerate}[label=(\roman*)]
\item Write $u _ { n }$ and $u _ { 2 n }$ in terms of $n$.
\item Hence show that $\log _ { 10 } \left( u _ { n } \right) - \log _ { 10 } \left( u _ { 2 n } \right) = n \log _ { 10 } \left( \frac { 4 } { 3 } \right)$.
\item Hence show that the value of
$$\log _ { 10 } \left( \frac { u _ { 100 } } { u _ { 200 } } \right)$$
is 12.5 correct to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q5 [12]}}