| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| 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 geometric series question testing standard formulas and basic logarithm manipulation. Part (a) uses the sum to infinity formula directly, part (b) is simple substitution and simplification, and part (c) involves writing the general term and applying basic log laws—all routine C2 techniques with no problem-solving or insight required. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{S_\infty =\}\ \dfrac{a}{1-r} = \dfrac{12}{1 - \frac{3}{8}}\) | M1 | \(\dfrac{a}{1-r}\) used |
| \(\{S_\infty =\}\ 19.2\) | A1 (Total: 2) | 19.2 OE; NMS mark as 2/2 or 0/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\)6th term \(=\}\ ar^{6-1}\) | M1 | Stated or used |
| \(= 12 \times \left(\frac{3}{8}\right)^5 = 2 \times 2 \times 3 \times \left(\frac{3}{2 \times 2 \times 2}\right)^5\) | m1 | Changing 8 and 12 in correct expression to correct products/powers of 2 and 3 |
| \(= \dfrac{2 \times 2 \times 3 \times 3^5}{(2^3)^5} = \dfrac{2^2 \times 3^6}{2^{15}} = \dfrac{3^6}{2^{13}}\) | A1 (Total: 3) | AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{u_n =\}\ 12 \times \left(\frac{3}{8}\right)^{n-1}\) | B1 (Total: 1) | OE e.g. \(32(3/8)^n\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log u_n = \log 12 + \log\left(\frac{3}{8}\right)^{n-1}\) | M1 | Using (c)(i) and taking logs: one log law used correctly, on a correct expression for \(u_n\) |
| \(\log u_n = \log 12 + (n-1)\log\left(\frac{3}{8}\right)\) | M1 | A second different log law used correctly, independent of previous M error and ft on c's (c)(i) provided c's \(u_n\) expression has a power involving \(n\) |
| \(\log u_n = \log 12 + (n-1)[\log 3 - \log 8]\) | m1 | A third different log law used correctly (or equivalent valid step) to reach a correct RHS whose terms are all multiples of \(\log 2\) and \(\log 3\). Dependent on both previous M marks |
| \(\log u_n = n\log 3 - 3n\log 2 + 5\log 2\) | ||
| \(\log_a u_n = n\log_a 3 - (3n-5)\log_a 2\) | A1 (Total: 4) | CSO AG Be convinced, no slips; condone absence of base \(a\) even in final line |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{S_\infty =\}\ \dfrac{a}{1-r} = \dfrac{12}{1 - \frac{3}{8}}$ | M1 | $\dfrac{a}{1-r}$ used |
| $\{S_\infty =\}\ 19.2$ | A1 (Total: 2) | 19.2 OE; NMS mark as 2/2 or 0/2 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{$6th term $=\}\ ar^{6-1}$ | M1 | Stated or used |
| $= 12 \times \left(\frac{3}{8}\right)^5 = 2 \times 2 \times 3 \times \left(\frac{3}{2 \times 2 \times 2}\right)^5$ | m1 | Changing 8 and 12 in correct expression to correct products/powers of 2 and 3 |
| $= \dfrac{2 \times 2 \times 3 \times 3^5}{(2^3)^5} = \dfrac{2^2 \times 3^6}{2^{15}} = \dfrac{3^6}{2^{13}}$ | A1 (Total: 3) | AG Be convinced |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{u_n =\}\ 12 \times \left(\frac{3}{8}\right)^{n-1}$ | B1 (Total: 1) | OE e.g. $32(3/8)^n$ |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log u_n = \log 12 + \log\left(\frac{3}{8}\right)^{n-1}$ | M1 | Using (c)(i) and taking logs: one log law used correctly, on a correct expression for $u_n$ |
| $\log u_n = \log 12 + (n-1)\log\left(\frac{3}{8}\right)$ | M1 | A second different log law used correctly, independent of previous M error and ft on c's (c)(i) provided c's $u_n$ expression has a power involving $n$ |
| $\log u_n = \log 12 + (n-1)[\log 3 - \log 8]$ | m1 | A third different log law used correctly (or equivalent valid step) to reach a correct RHS whose terms are all multiples of $\log 2$ and $\log 3$. Dependent on both previous M marks |
| $\log u_n = n\log 3 - 3n\log 2 + 5\log 2$ | | |
| $\log_a u_n = n\log_a 3 - (3n-5)\log_a 2$ | A1 (Total: 4) | CSO AG Be convinced, no slips; condone absence of base $a$ even in final line |9 The first term of a geometric series is 12 and the common ratio of the series is $\frac { 3 } { 8 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the sum to infinity of the series.
\item Show that the sixth term of the series can be written in the form $\frac { 3 ^ { 6 } } { 2 ^ { 13 } }$.
\item The $n$th term of the series is $u _ { n }$.
\begin{enumerate}[label=(\roman*)]
\item Write down an expression for $u _ { n }$ in terms of $n$.
\item Hence show that
$$\log _ { a } u _ { n } = n \log _ { a } 3 - ( 3 n - 5 ) \log _ { a } 2$$
(4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q9 [10]}}