| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Largest or extreme value of sum |
| Difficulty | Standard +0.3 This is a standard C2 arithmetic series question with routine algebraic manipulation. Part (a) applies the sum formula directly, part (b) solves simultaneous equations, and part (c) requires recognizing that the maximum sum occurs when terms change sign—a common textbook exercise. All techniques are straightforward with no novel insight required, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S_5 = \frac{5}{2}(2a + 4d) = 575\) | M1 | Correct use of sum formula |
| \(5(a + 2d) = 575\) | A1 | Correct simplification |
| \(a + 2d = 115\) | A1 | Shown/confirmed (given result) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 10th term: \(a + 9d = 87\) | M1 | Correct use of \(n\)th term formula |
| Solving simultaneously: \((a+9d)-(a+2d) = 87-115\), so \(7d = -28\) | M1 | Correct method to eliminate \(a\) |
| \(d = -4\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 115 - 2(-4) = 123\) | B1 | Correct value of \(a\) |
| \(u_n = 123 + (n-1)(-4) = 127 - 4n\) | M1 | Correct expression for \(u_n\) |
| \(u_k > 0 \Rightarrow 127 - 4k > 0 \Rightarrow k < 31.75\), and \(u_{k+1} < 0 \Rightarrow k+1 > 31.75\), so \(k = 31\) | M1 | Correct method to find \(k\) |
| \(\sum_{n=1}^{31} u_n = \frac{31}{2}(2 \times 123 + 30 \times (-4)) = \frac{31}{2}(246-120) = \frac{31}{2}(126)\) | M1 | Correct use of sum formula with \(k=31\) |
| \(= 1953\) | A1 | Correct final answer |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $S_5 = \frac{5}{2}(2a + 4d) = 575$ | M1 | Correct use of sum formula |
| $5(a + 2d) = 575$ | A1 | Correct simplification |
| $a + 2d = 115$ | A1 | Shown/confirmed (given result) |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| 10th term: $a + 9d = 87$ | M1 | Correct use of $n$th term formula |
| Solving simultaneously: $(a+9d)-(a+2d) = 87-115$, so $7d = -28$ | M1 | Correct method to eliminate $a$ |
| $d = -4$ | A1 | Correct value |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 115 - 2(-4) = 123$ | B1 | Correct value of $a$ |
| $u_n = 123 + (n-1)(-4) = 127 - 4n$ | M1 | Correct expression for $u_n$ |
| $u_k > 0 \Rightarrow 127 - 4k > 0 \Rightarrow k < 31.75$, and $u_{k+1} < 0 \Rightarrow k+1 > 31.75$, so $k = 31$ | M1 | Correct method to find $k$ |
| $\sum_{n=1}^{31} u_n = \frac{31}{2}(2 \times 123 + 30 \times (-4)) = \frac{31}{2}(246-120) = \frac{31}{2}(126)$ | M1 | Correct use of sum formula with $k=31$ |
| $= 1953$ | A1 | Correct final answer |
I can see these are answer space pages from an AQA MPC2 exam paper (June 2014), but the mark scheme content itself is not shown in these images - they only show blank answer spaces for Questions 8 and 9.
However, based on the question content visible on page 18, I can work through the solutions:
---8 An arithmetic series has first term $a$ and common difference $d$. The sum of the first 5 terms of the series is 575 .
\begin{enumerate}[label=(\alph*)]
\item Show that $a + 2 d = 115$.
\item Given also that the 10th term of the series is 87, find the value of $d$.
\item The $n$th term of the series is $u _ { n }$. Given that $u _ { k } > 0$ and $u _ { k + 1 } < 0$, find the value of $\sum _ { n = 1 } ^ { k } u _ { n }$.\\[0pt]
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2014 Q8 [11]}}