| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find parameter from given term |
| Difficulty | Moderate -0.3 This is a straightforward recurrence relation question requiring substitution to find a₂ and a₃, followed by solving a simple equation. Part (a) involves basic algebraic manipulation, and part (b) requires solving a quadratic equation. While it has multiple steps, each step is routine and the question follows a standard C1 pattern with no novel insight required. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(a_2 = 2k\) | B1 | \(2k\) only |
| \(a_3 = \frac{k(\text{"\)2k\("}+1)}{\text{"\)2k\("}}\) | M1 | For substituting their \(a_2\) into \(a_3 = \frac{k(a_2+1)}{a_2}\) to find \(a_3\) in terms of \(k\) only |
| \(a_3 = \frac{2k+1}{2}\) | A1 | Or exact simplified equivalent such as \(k + \frac{1}{2}\) or \(\frac{1}{2}(2k+1)\) but not \(k + \frac{k}{2k}\). Must be seen in (a) but isw once correct simplified answer seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sum_{r=1}^{3} a_r = 10 \Rightarrow 1 + \text{"\)2k\("} + \text{"\)\frac{2k+1}{2}\("} = 10\) | M1 | Writes \(1 +\) their \(a_2 +\) their \(a_3 = 10\). E.g. \(1 + 2k + \frac{2k^2+k}{2k} = 10\). Must be correct follow-through equation in terms of \(k\) only |
| \(\Rightarrow 2 + 4k + 2k + 1 = 20 \Rightarrow k = \ldots\) or e.g. \(\Rightarrow 6k^2 - 17k = 0 \Rightarrow k = \ldots\) | M1 | Solves their equation in \(k\) from sum of 3 terms \(= 10\). Condone poor algebra but if quadratic obtained then usual rules apply for solving |
| \(k = \frac{17}{6}\) | A1 | Or exact equivalent e.g. \(2\frac{5}{6}\). Do not allow \(k = \frac{8.5}{3}\) or \(k = \frac{17/2}{3}\). Ignore reference to \(k=0\). Allow \(2.8\overline{3}\) if recurring clearly indicated |
## Question 3(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $a_2 = 2k$ | B1 | $2k$ only |
| $a_3 = \frac{k(\text{"$2k$"}+1)}{\text{"$2k$"}}$ | M1 | For substituting their $a_2$ into $a_3 = \frac{k(a_2+1)}{a_2}$ to find $a_3$ in terms of $k$ only |
| $a_3 = \frac{2k+1}{2}$ | A1 | Or exact simplified equivalent such as $k + \frac{1}{2}$ or $\frac{1}{2}(2k+1)$ but not $k + \frac{k}{2k}$. **Must be seen in (a)** but isw once correct simplified answer seen |
**Subtotal: 3 marks**
---
## Question 3(b):
*Note: no marks for using an AP (or GP) sum formula unless terms form an AP (or GP).*
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{3} a_r = 10 \Rightarrow 1 + \text{"$2k$"} + \text{"$\frac{2k+1}{2}$"} = 10$ | M1 | Writes $1 +$ their $a_2 +$ their $a_3 = 10$. E.g. $1 + 2k + \frac{2k^2+k}{2k} = 10$. Must be correct follow-through equation in terms of $k$ only |
| $\Rightarrow 2 + 4k + 2k + 1 = 20 \Rightarrow k = \ldots$ or e.g. $\Rightarrow 6k^2 - 17k = 0 \Rightarrow k = \ldots$ | M1 | Solves their equation in $k$ from sum of 3 terms $= 10$. Condone poor algebra but if quadratic obtained then usual rules apply for solving |
| $k = \frac{17}{6}$ | A1 | Or exact equivalent e.g. $2\frac{5}{6}$. Do **not** allow $k = \frac{8.5}{3}$ or $k = \frac{17/2}{3}$. Ignore reference to $k=0$. Allow $2.8\overline{3}$ if recurring clearly indicated |
**Subtotal: 3 marks**
**Total: 6 marks**\begin{enumerate}
\item A sequence $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ is defined by
\end{enumerate}
$$\begin{aligned}
a _ { 1 } & = 1 \\
a _ { n + 1 } & = \frac { k \left( a _ { n } + 1 \right) } { a _ { n } } , \quad n \geqslant 1
\end{aligned}$$
where $k$ is a positive constant.\\
(a) Write down expressions for $a _ { 2 }$ and $a _ { 3 }$ in terms of $k$, giving your answers in their simplest form.
Given that $\sum _ { r = 1 } ^ { 3 } a _ { r } = 10$\\
(b) find an exact value for $k$.\\
\hfill \mbox{\textit{Edexcel C1 2017 Q3 [6]}}