| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find parameter from given term |
| Difficulty | Standard +0.3 This is a straightforward recurrence relation question requiring substitution to find a₂ and a₃, then solving a quadratic equation. While it involves multiple steps, the techniques are routine for C1 level—direct substitution and basic algebra—making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a_2 = k(4+2) \ (= 6k)\) | B1 | Any correct (possibly un-simplified) expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a_3 = k(\text{their } a_2 + 2) \ (= 6k^2 + 2k)\) | M1 | An attempt at \(a_3\). Can follow through their answer to (a) but \(a_2\) must be an expression in \(k\). |
| \(a_1 + a_2 + a_3 = 4 + (6k) + (6k^2 + 2k)\) | M1 | An attempt to find their \(a_1 + a_2 + a_3\) |
| \(4 + (6k) + (6k^2 + 2k) = 2\) | A1 | A correct equation in any form |
| Solves \(6k^2 + 8k + 2 = 0\) to obtain \(k=\ldots\) \((6k^2 + 8k + 2 = 2(3k+1)(k+1))\) | M1 | Solves their 3TQ as far as \(k = \ldots\) according to general principles. (An independent mark for solving their three term quadratic) |
| \(k = -\frac{1}{3}\) | A1 | Any equivalent fraction |
| \(k = -1\) | B1 | Must be from a correct equation. (Do not accept un-simplified) |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a_2 = k(4+2) \ (= 6k)$ | B1 | Any correct (possibly un-simplified) expression |
**Total: (1)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a_3 = k(\text{their } a_2 + 2) \ (= 6k^2 + 2k)$ | M1 | An attempt at $a_3$. Can follow through their answer to (a) but $a_2$ must be an expression in $k$. |
| $a_1 + a_2 + a_3 = 4 + (6k) + (6k^2 + 2k)$ | M1 | An attempt to find their $a_1 + a_2 + a_3$ |
| $4 + (6k) + (6k^2 + 2k) = 2$ | A1 | A **correct** equation in any form |
| Solves $6k^2 + 8k + 2 = 0$ to obtain $k=\ldots$ $(6k^2 + 8k + 2 = 2(3k+1)(k+1))$ | M1 | Solves their 3TQ as far as $k = \ldots$ according to general principles. (An independent mark for solving their three term quadratic) |
| $k = -\frac{1}{3}$ | A1 | Any equivalent fraction |
| $k = -1$ | B1 | Must be from a correct equation. (Do not accept un-simplified) |
**Total: (6) [7]**
---4. A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
a _ { 1 } & = 4 \\
a _ { n + 1 } & = k \left( a _ { n } + 2 \right) , \quad \text { for } n \geqslant 1
\end{aligned}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $a _ { 2 }$ in terms of $k$.
Given that $\sum _ { i = 1 } ^ { 3 } a _ { i } = 2$,
\item find the two possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q4 [7]}}