| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find n given sum condition |
| Difficulty | Standard +0.3 Part (a) is a straightforward 'show that' using the standard sum formula S_n = n/2[2a + (n-1)d]. Part (b) requires solving two simultaneous equations using basic algebra. Part (c) involves algebraic manipulation of the given equation to find a sum, but once S_25 = 3500 is substituted, it becomes a simple linear equation. All steps are routine applications of standard formulas with no novel insight required, making this slightly easier than average. |
| Spec | 1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(S_{25} = \frac{25}{2}[2a + (25-1)d]\) | M1 | \(\frac{25}{2}[2a+(25-1)d]\) OE |
| \(\frac{25}{2}[2a+24d]=3500\) | ||
| \(25(2a+24d)=7000\) or \(\frac{50a+600d}{2}=3500\) | m1 | Forming equation and attempt to remove fraction or expand brackets |
| \(50a+600d=7000\), so \(a+12d=140\) | A1 | CSO AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(5^{\text{th}}\) term \(= a+4d\) | M1 | \(a+(5-1)d\) used correctly |
| \(a+12d=140\), \(a+4d=100\), \(\Rightarrow 8d=40\) | M1 | Solving \(a+12d=140\) simultaneously with either \(a+4d=100\) or \(a+5d=100\), eliminating \(a\) or \(d\) |
| \(d=5\) | A1 | |
| \(a=80\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(33\left(3500-\sum_{n=1}^{k}u_n\right)=67\sum_{n=1}^{k}u_n\) | M1 | Recognition that \(\sum_{n=1}^{25}u_n=3500\) |
| \(33\times3500=67\sum_{n=1}^{k}u_n+33\sum_{n=1}^{k}u_n\) | m1 | Correct rearrangement PI |
| \(100\times\sum_{n=1}^{k}u_n=33\times3500 \Rightarrow \sum_{n=1}^{k}u_n=1155\) | A1 |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $S_{25} = \frac{25}{2}[2a + (25-1)d]$ | M1 | $\frac{25}{2}[2a+(25-1)d]$ OE |
| $\frac{25}{2}[2a+24d]=3500$ | | |
| $25(2a+24d)=7000$ or $\frac{50a+600d}{2}=3500$ | m1 | Forming equation and attempt to remove fraction or expand brackets |
| $50a+600d=7000$, so $a+12d=140$ | A1 | CSO AG Be convinced |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $5^{\text{th}}$ term $= a+4d$ | M1 | $a+(5-1)d$ used correctly |
| $a+12d=140$, $a+4d=100$, $\Rightarrow 8d=40$ | M1 | Solving $a+12d=140$ simultaneously with either $a+4d=100$ or $a+5d=100$, eliminating $a$ or $d$ |
| $d=5$ | A1 | |
| $a=80$ | A1 | |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $33\left(3500-\sum_{n=1}^{k}u_n\right)=67\sum_{n=1}^{k}u_n$ | M1 | Recognition that $\sum_{n=1}^{25}u_n=3500$ |
| $33\times3500=67\sum_{n=1}^{k}u_n+33\sum_{n=1}^{k}u_n$ | m1 | Correct rearrangement PI |
| $100\times\sum_{n=1}^{k}u_n=33\times3500 \Rightarrow \sum_{n=1}^{k}u_n=1155$ | A1 | |
---6 An arithmetic series has first term $a$ and common difference $d$.
The sum of the first 25 terms of the series is 3500 .
\begin{enumerate}[label=(\alph*)]
\item Show that $a + 12 d = 140$.
\item The fifth term of this series is 100 .
Find the value of $d$ and the value of $a$.
\item The $n$th term of this series is $u _ { n }$. Given that
$$33 \left( \sum _ { n = 1 } ^ { 25 } u _ { n } - \sum _ { n = 1 } ^ { k } u _ { n } \right) = 67 \sum _ { n = 1 } ^ { k } u _ { n }$$
find the value of $\sum _ { n = 1 } ^ { k } u _ { n }$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q6 [10]}}