| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Arithmetic progression with parameters |
| Difficulty | Moderate -0.5 This is a straightforward application of standard arithmetic series formulas (S_n and nth term). Part (a) requires substituting n=9 into the sum formula and simplifying algebraically. Part (b) involves setting up a second equation from the given condition and solving simultaneous equations. While it requires multiple steps, all techniques are routine and commonly practiced in C1/C2 with no novel insight needed, making it slightly easier than average. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_9 = 54 \Rightarrow 54 = \frac{9}{2}(2a + 8d)\) or \(54 = \frac{9}{2}(a + a + 8d)\) | M1 | Uses a correct sum formula with \(n=9\) and \(S_9 = 54\) |
| \(a + 4d = 6\) | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a + 7d = \frac{1}{2}(a + 6d)\) or \(\frac{1}{2}(a+7d) = a + 6d\) | M1 | Uses \(t_8 = \frac{1}{2}t_7\) or \(\frac{1}{2}t_8 = t_7\) to produce one of these equations |
| \(6 - 4d + 7d = \frac{1}{2}(6 - 4d + 6d) \Rightarrow d = \ldots\) | M1 | Uses the given equation from (a) and their second linear equation in \(a\) and \(d\), proceeds to find a value for either \(a\) or \(d\) |
| \(d = -1.5, \ a = 12\) | A1A1 | A1: Either \(d = -1.5\) (oe) or \(a = 12\); A1: Both \(d = -1.5\) (oe) and \(a = 12\) |
## Question 4:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_9 = 54 \Rightarrow 54 = \frac{9}{2}(2a + 8d)$ or $54 = \frac{9}{2}(a + a + 8d)$ | M1 | Uses a correct sum formula with $n=9$ and $S_9 = 54$ |
| $a + 4d = 6$ | A1* | cso |
**Listing method:** $a+a+d+a+2d+\ldots+a+8d = 54 \Rightarrow 9a + 36d = 54$. Scores M1 for attempting to sum 9 terms (both lines needed).
---
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a + 7d = \frac{1}{2}(a + 6d)$ or $\frac{1}{2}(a+7d) = a + 6d$ | M1 | Uses $t_8 = \frac{1}{2}t_7$ or $\frac{1}{2}t_8 = t_7$ to produce one of these equations |
| $6 - 4d + 7d = \frac{1}{2}(6 - 4d + 6d) \Rightarrow d = \ldots$ | M1 | Uses the given equation from (a) and their second linear equation in $a$ and $d$, proceeds to find a value for either $a$ or $d$ |
| $d = -1.5, \ a = 12$ | A1A1 | A1: Either $d = -1.5$ (oe) or $a = 12$; A1: Both $d = -1.5$ (oe) and $a = 12$ |
**Note:** Use of $\frac{1}{2}t_8 = t_7$ in (b) gives $a = 30$ and $d = -6$.
---4. An arithmetic series has first term $a$ and common difference $d$.
Given that the sum of the first 9 terms is 54
\begin{enumerate}[label=(\alph*)]
\item show that
$$a + 4 d = 6$$
Given also that the 8th term is half the 7th term,
\item find the values of $a$ and $d$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q4 [6]}}