| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Find n given sum condition |
| Difficulty | Standard +0.3 This is a straightforward application of standard arithmetic sequence formulas (a_n = a + (n-1)d and S_n = n/2(2a + (n-1)d)). Part (a) involves solving simultaneous equations with clear guidance ('Show that'), and part (b) requires solving a linear inequality. While it requires multiple steps, all techniques are routine for C2 level with no novel problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| \((S_{10} =) \frac{40}{2}[2a + (40-1)d]\) | M1 | |
| \(20(2a + 39d) = 1250\) | A1 | |
| \((25\text{th term}) = a + (25-1)d\) | M1 | |
| \(a + 24d = 38\) | A1 | |
| m1 | Dep on both previous two Ms. Solving two equations in \(a\) and \(d\) simultaneously. | |
| \(18d = 27 \Rightarrow d = 1.5\) | A1cso | AG Be convinced. SC Using the given answer for \(d\): mark out of a maximum of 4/6 as M1A1M1A1 (conclusion also needed in last A mark) (m0A0). 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 38 - 24 \times 1.5 = 2\) | M1 | PI if using \(a = 2\) in (b). If using eg \(a = 38\) award this M mark at \(100 - 38 + 1 + 24\) stage: no. of terms \(\frac{100-38}{1.5} + 1 + 24\) |
| \(a + (n-1)1.5 < 100\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \text{number of terms} < 100 \text{ is } 66\) | A1 | NMS mark as B3 for 66 else B0. 3 marks total |
**8(a)**
$(S_{10} =) \frac{40}{2}[2a + (40-1)d]$ | M1 |
$20(2a + 39d) = 1250$ | A1 |
$(25\text{th term}) = a + (25-1)d$ | M1 |
$a + 24d = 38$ | A1 |
| m1 | Dep on both previous two Ms. Solving two equations in $a$ and $d$ simultaneously.
$18d = 27 \Rightarrow d = 1.5$ | A1cso | AG Be convinced. SC Using the given answer for $d$: mark out of a maximum of 4/6 as M1A1M1A1 (conclusion also needed in last A mark) (m0A0). 6 marks total
**8(b)**
$a = 38 - 24 \times 1.5 = 2$ | M1 | PI if using $a = 2$ in (b). If using eg $a = 38$ award this M mark at $100 - 38 + 1 + 24$ stage: no. of terms $\frac{100-38}{1.5} + 1 + 24$
$a + (n-1)1.5 < 100$ | M1 |
$n < \frac{100-a}{1.5} + 1$
$n < 66.333\ldots$
$\Rightarrow \text{number of terms} < 100 \text{ is } 66$ | A1 | NMS mark as B3 for 66 else B0. 3 marks total
**Total for Q8: 9 marks**
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## **GRAND TOTAL: 75 marks**8 The 25th term of an arithmetic series is 38 .\\
The sum of the first 40 terms of the series is 1250 .
\begin{enumerate}[label=(\alph*)]
\item Show that the common difference of this series is 1.5 .
\item Find the number of terms in the series which are less than 100 .
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2009 Q8 [9]}}