| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Linear iterative formula u(n+1) = pu(n) + q |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring only direct substitution for part (a) and solving a simple linear equation L = 6 + (2/5)L for part (b). Both parts are routine C2-level exercises with no problem-solving insight needed, making it easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_2 = 6 + \frac{2}{5}(2) = 6.8\) | B1 | |
| \(u_3 = 6 + \frac{2}{5}(6.8) = 8.72\) | B1 | ft from \(u_2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(L = 6 + \frac{2}{5}L\) | M1 | Setting \(u_{n+1} = u_n = L\) |
| \(\frac{3}{5}L = 6\) | M1 | Rearranging |
| \(L = 10\) | A1 | cao |
# Question 2:
## Part (a)
| $u_2 = 6 + \frac{2}{5}(2) = 6.8$ | B1 | |
|---|---|---|
| $u_3 = 6 + \frac{2}{5}(6.8) = 8.72$ | B1 | ft from $u_2$ |
## Part (b)
| $L = 6 + \frac{2}{5}L$ | M1 | Setting $u_{n+1} = u_n = L$ |
|---|---|---|
| $\frac{3}{5}L = 6$ | M1 | Rearranging |
| $L = 10$ | A1 | cao |
---2 The $n$th term of a sequence is $u _ { n }$.\\
The sequence is defined by
$$u _ { n + 1 } = 6 + \frac { 2 } { 5 } u _ { n }$$
The first term of the sequence is given by $u _ { 1 } = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $u _ { 2 }$ and the value of $u _ { 3 }$.
\item The limit of $u _ { n }$ as $n$ tends to infinity is $L$.
Write down an equation for $L$ and hence find the value of $L$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q2 [5]}}