| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Shared terms between AP and GP |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard formulas for GP sum to infinity (S = a/(1-r)) and AP properties. Part (a) involves simple algebraic manipulation, while part (b) requires setting up equations linking the two sequences and applying the arithmetic series sum formula. All techniques are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_\infty = \frac{a}{1-r}\), so \(50 = \frac{10}{1-r}\) | M1 | Use of correct formula |
| \(1-r = \frac{10}{50} = \frac{1}{5}\), hence \(r = \frac{4}{5}\) | A1 | Shown convincingly |
| Answer | Marks |
|---|---|
| Second term \(= 10 \times \frac{4}{5} = 8\) | B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4th term of AP \(= 10\), 8th term of AP \(= 8\) | ||
| \(a + 3d = 10\) and \(a + 7d = 8\) | M1 | Setting up two equations |
| \(4d = -2\), so \(d = -\frac{1}{2}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a + 3(-\frac{1}{2}) = 10 \Rightarrow a = 11.5\) | M1 | Finding first term |
| \(\sum_{n=1}^{40} u_n = \frac{40}{2}(2(11.5) + 39(-\frac{1}{2}))\) | M1 | Use of correct sum formula |
| \(= 20(23 - 19.5) = 20 \times 3.5\) | A1 | |
| \(= 70\) | A1 | cao |
# Question 5:
## Part (a)(i)
| $S_\infty = \frac{a}{1-r}$, so $50 = \frac{10}{1-r}$ | M1 | Use of correct formula |
|---|---|---|
| $1-r = \frac{10}{50} = \frac{1}{5}$, hence $r = \frac{4}{5}$ | A1 | Shown convincingly |
## Part (a)(ii)
| Second term $= 10 \times \frac{4}{5} = 8$ | B1 B1 | |
## Part (b)(i)
| 4th term of AP $= 10$, 8th term of AP $= 8$ | | |
|---|---|---|
| $a + 3d = 10$ and $a + 7d = 8$ | M1 | Setting up two equations |
| $4d = -2$, so $d = -\frac{1}{2}$ | A1 A1 | |
## Part (b)(ii)
| $a + 3(-\frac{1}{2}) = 10 \Rightarrow a = 11.5$ | M1 | Finding first term |
|---|---|---|
| $\sum_{n=1}^{40} u_n = \frac{40}{2}(2(11.5) + 39(-\frac{1}{2}))$ | M1 | Use of correct sum formula |
| $= 20(23 - 19.5) = 20 \times 3.5$ | A1 | |
| $= 70$ | A1 | cao |5
\begin{enumerate}[label=(\alph*)]
\item An infinite geometric series has common ratio $r$.\\
The first term of the series is 10 and its sum to infinity is 50 .
\begin{enumerate}[label=(\roman*)]
\item Show that $r = \frac { 4 } { 5 }$.
\item Find the second term of the series.
\end{enumerate}\item The first and second terms of the geometric series in part (a) have the same values as the 4th and 8th terms respectively of an arithmetic series.
\begin{enumerate}[label=(\roman*)]
\item Find the common difference of the arithmetic series.
\item The $n$th term of the arithmetic series is $u _ { n }$. Find the value of $\sum _ { n = 1 } ^ { 40 } u _ { n }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q5 [11]}}