| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| 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 | Standard +0.3 This is a straightforward recurrence relation problem requiring substitution and algebraic manipulation. Students substitute u₁ into the formula to get u₂, then substitute u₂ to get u₃ in terms of k, and finally solve u₃ = 6 to find k. While it involves some fractional algebra, it's a standard multi-step exercise with clear methodology and no novel insight required—slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_2 = k - \frac{8}{1}\) or \(u_3 = k - \frac{8}{"k-8"}\) | M1 | Either expression attempted |
| \(u_2 = k - \frac{8}{1}\) and \(u_3 = k - \frac{8}{k-8}\) | A1 | Both correct (isw after correct unsimplified expression for both) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_3 = 6 \Rightarrow k - \frac{8}{k-8} = 6\) | M1 | Sets \(u_3 = 6\) to form an equation in \(k\) |
| \(k^2 - 14k + 40 = 0\) | M1 | Rearranges to form a 3TQ = 0 or equivalent; may be implied by further work |
| \((k-4)(k-10) = 0 \Rightarrow k = \ldots\) or \(k = \frac{14 \pm \sqrt{14^2 - 4\times1\times40}}{2\times1}\) | dM1 | Attempts to solve their 3TQ; dependent on previous M mark |
| \(k = 4, \ 10\) | A1 | Both values; do not allow \(x = \ldots\) |
## Question 5:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_2 = k - \frac{8}{1}$ or $u_3 = k - \frac{8}{"k-8"}$ | M1 | Either expression attempted |
| $u_2 = k - \frac{8}{1}$ and $u_3 = k - \frac{8}{k-8}$ | A1 | Both correct (isw after correct unsimplified expression for both) |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_3 = 6 \Rightarrow k - \frac{8}{k-8} = 6$ | M1 | Sets $u_3 = 6$ to form an equation in $k$ |
| $k^2 - 14k + 40 = 0$ | M1 | Rearranges to form a 3TQ = 0 or equivalent; may be implied by further work |
| $(k-4)(k-10) = 0 \Rightarrow k = \ldots$ or $k = \frac{14 \pm \sqrt{14^2 - 4\times1\times40}}{2\times1}$ | dM1 | Attempts to solve their 3TQ; dependent on previous M mark |
| $k = 4, \ 10$ | A1 | Both values; do not allow $x = \ldots$ |
---5. A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
$$\begin{aligned}
u _ { 1 } & = 1 \\
u _ { n + 1 } & = k - \frac { 8 } { u _ { n } } \quad n \geqslant 1
\end{aligned}$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Write down expressions for $u _ { 2 }$ and $u _ { 3 }$ in terms of $k$.
Given that $u _ { 3 } = 6$
\item find the possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q5 [6]}}