| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Moderate -0.8 This is a straightforward geometric series question requiring standard formulas and techniques. Part (a) uses the ratio of terms formula (u₆/u₃ = r³), part (b)(i) applies the sum formula directly, and part (b)(ii) involves solving an exponential inequality with logarithms. All steps are routine C2-level procedures with no novel problem-solving required, making it easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_3 = ar^2 = 36\) and \(u_6 = ar^5 = 972\) | M1 | Setting up two equations using geometric series formula |
| \(\frac{ar^5}{ar^2} = r^3 = \frac{972}{36} = 27\), so \(r = 3\) | A1 | Dividing equations and reaching \(r=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a \times 9 = 36\) | M1 | Substituting \(r=3\) into \(ar^2=36\) |
| \(a = 4\) | A1 | Correct first term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{20} = \frac{4(3^{20}-1)}{3-1}\) | M1 | Correct sum formula with \(a=4\), \(r=3\), \(n=20\) |
| \(= \frac{4(3^{20}-1)}{2} = 2(3^{20}-1)\) | A1 | Simplification to given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4 \times 3^{n-1} > 4\times10^{15}\) | M1 | Setting up inequality with \(u_n = 4\times3^{n-1}\) |
| \(3^{n-1} > 10^{15}\), \((n-1)\log 3 > 15\log 10\) | M1 | Taking logarithms |
| \(n-1 > \frac{15}{\log 3} = 31.4...\), so \(n-1 \geq 32\), \(n \geq 33\) | A1 | \(n = 33\) |
# Question 6:
## Part (a)(i) – Show common ratio = 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_3 = ar^2 = 36$ and $u_6 = ar^5 = 972$ | M1 | Setting up two equations using geometric series formula |
| $\frac{ar^5}{ar^2} = r^3 = \frac{972}{36} = 27$, so $r = 3$ | A1 | Dividing equations and reaching $r=3$ |
## Part (a)(ii) – Find first term
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a \times 9 = 36$ | M1 | Substituting $r=3$ into $ar^2=36$ |
| $a = 4$ | A1 | Correct first term |
## Part (b)(i) – Show sum = $2(3^{20}-1)$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{20} = \frac{4(3^{20}-1)}{3-1}$ | M1 | Correct sum formula with $a=4$, $r=3$, $n=20$ |
| $= \frac{4(3^{20}-1)}{2} = 2(3^{20}-1)$ | A1 | Simplification to given result |
## Part (b)(ii) – Least value of $n$ such that $u_n > 4\times10^{15}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 \times 3^{n-1} > 4\times10^{15}$ | M1 | Setting up inequality with $u_n = 4\times3^{n-1}$ |
| $3^{n-1} > 10^{15}$, $(n-1)\log 3 > 15\log 10$ | M1 | Taking logarithms |
| $n-1 > \frac{15}{\log 3} = 31.4...$, so $n-1 \geq 32$, $n \geq 33$ | A1 | $n = 33$ |
---6 A geometric series has third term 36 and sixth term 972.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the common ratio of the series is 3 .
\item Find the first term of the series.
\end{enumerate}\item The $n$th term of the series is $u _ { n }$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\sum _ { n = 1 } ^ { 20 } u _ { n } = 2 \left( 3 ^ { 20 } - 1 \right)$.
\item Find the least value of $n$ such that $u _ { n } > 4 \times 10 ^ { 15 }$.\\
$7 \quad$ A curve $C$ is defined for $x > 0$ by the equation $y = x + 3 + \frac { 8 } { x ^ { 4 } }$ and is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{1c06ba04-575c-4eb8-b4aa-0a7510838cd2-08_602_799_447_632}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q6 [9]}}