| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find n given sum condition |
| Difficulty | Moderate -0.3 This is a straightforward multi-part arithmetic series question requiring standard formula application (S_n = n/2[2a + (n-1)d]) and algebraic manipulation. Part (a) is routine verification, part (b) involves solving simultaneous equations from given conditions, and part (c) requires finding k when the sum equals zero. All steps follow predictable patterns with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S_{31} = \frac{31}{2}[2a + (31-1)d]\) | M1 | |
| \(31(a + 15d) = 310\) | m1 | Forming eqn and eliminating fraction or bracket |
| \(a + 15d = 310/31;\ a + 15d = 10\) | A1 | AG Completion to printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a + (21-1)d = 2[a+(16-1)d]\) | M1 | \(a+(n-1)d\) used for at least one term |
| \(\Rightarrow a = -10d;\ \Rightarrow -10d + 15d = 10\) | m1 | Solving \(a+15d=10\) simultaneously with eqn in \(a\) and \(d\) obtained from \(a+20d = k[a+15d]\) with \(k=2\) or \(k=\frac{1}{2}\) |
| \(d = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u_1 = a = -20\) | B1F | ft on c's value for \(d\) in \(a+15d=10\) or in another correct equation in \(a\) and \(d\); value for \(a\) must appear within c's solution for (c) |
| \(\sum_{n=1}^{k} u_n = S_k = \frac{k}{2}[2a+(k-1)d]\) | M1 | Condone \(n\) for \(k\) in M1 and A1F lines provided \(n\) replaced by \(k\) at later stage |
| \(\frac{k}{2}[-40+2k-2] = 0\) | A1F | \(= 0\) can be implied by later line; ft on c's non-zero values for \(a\) and \(d\) |
| \(k = 21\) | A1 | Condone presence of \(k=0\). SC NMS \(k=21\) and \(d=2\) found earlier award B2. If \(k=21\) but never see \(d=2\), award 0/4 |
## Question 4:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{31} = \frac{31}{2}[2a + (31-1)d]$ | M1 | |
| $31(a + 15d) = 310$ | m1 | Forming eqn and eliminating fraction or bracket |
| $a + 15d = 310/31;\ a + 15d = 10$ | A1 | AG Completion to printed answer |
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a + (21-1)d = 2[a+(16-1)d]$ | M1 | $a+(n-1)d$ used for at least one term |
| $\Rightarrow a = -10d;\ \Rightarrow -10d + 15d = 10$ | m1 | Solving $a+15d=10$ simultaneously with eqn in $a$ and $d$ obtained from $a+20d = k[a+15d]$ with $k=2$ or $k=\frac{1}{2}$ |
| $d = 2$ | A1 | |
**Part (c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u_1 = a = -20$ | B1F | ft on c's value for $d$ in $a+15d=10$ or in another correct equation in $a$ and $d$; value for $a$ must appear within c's solution for (c) |
| $\sum_{n=1}^{k} u_n = S_k = \frac{k}{2}[2a+(k-1)d]$ | M1 | Condone $n$ for $k$ in M1 and A1F lines provided $n$ replaced by $k$ at later stage |
| $\frac{k}{2}[-40+2k-2] = 0$ | A1F | $= 0$ can be implied by later line; ft on c's non-zero values for $a$ and $d$ |
| $k = 21$ | A1 | Condone presence of $k=0$. **SC** NMS $k=21$ and $d=2$ found earlier award B2. If $k=21$ but never see $d=2$, award 0/4 |
---4 An arithmetic series has first term $a$ and common difference $d$.\\
The sum of the first 31 terms of the series is 310 .
\begin{enumerate}[label=(\alph*)]
\item Show that $a + 15 d = 10$.
\item Given also that the 21st term is twice the 16th term, find the value of $d$.
\item The $n$th term of the series is $u _ { n }$. Given that $\sum _ { n = 1 } ^ { k } u _ { n } = 0$, find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q4 [10]}}