| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| 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 -0.3 This is a straightforward recurrence relation question requiring substitution to find constants, then basic iteration and limit concepts. Part (a) involves solving simultaneous equations from given terms, part (b) is direct substitution, and part (c) uses the standard result that if u_{n+1} = pu_n + q converges to L, then L = pL + q. All techniques are routine for C2 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(8 = -8p + q\); \(4 = 8p + q\) | M1, A1 | Either equation. PI eg by combined eqn. Both (condone embedded values for the M1A1) |
| Valid method to solve two simultaneous equations in \(p\) and \(q\) to find either \(p\) or \(q\) | m1 (5 marks) | Valid method to solve two simultaneous equations in \(p\) and \(q\) to find either \(p\) or \(q\) |
| \(q = 6\); \(p = -0.25\) | A1, B1 | AG (condone if left as a fraction); OE |
| (b) \(u_4 = 5\) | B1F (1 mark) | Ft on (6 + 4p) |
| (c)(i) \(L = pL + q\); \((L = -0.25L + 6)\) | M1 (1 mark) | OE |
| (ii) \(L = \frac{q}{1-p}\) | m1 | Rearranging |
| \(L = \frac{6}{1.25} = 4.8\) | A1F (2 marks) | Ft on \(\frac{6}{1-p}\); Dependent on previous two marks |
**(a)** $8 = -8p + q$; $4 = 8p + q$ | M1, A1 | Either equation. PI eg by combined eqn. Both (condone embedded values for the M1A1)
Valid method to solve two simultaneous equations in $p$ and $q$ to find either $p$ or $q$ | m1 (5 marks) | Valid method to solve two simultaneous equations in $p$ and $q$ to find either $p$ or $q$
$q = 6$; $p = -0.25$ | A1, B1 | AG (condone if left as a fraction); OE
**(b)** $u_4 = 5$ | B1F (1 mark) | Ft on (6 + 4p)
**(c)(i)** $L = pL + q$; $(L = -0.25L + 6)$ | M1 (1 mark) | OE
**(ii)** $L = \frac{q}{1-p}$ | m1 | Rearranging
$L = \frac{6}{1.25} = 4.8$ | A1F (2 marks) | Ft on $\frac{6}{1-p}$; Dependent on previous two marks
**Total for Q6: 9 marks**
---6 The $n$th term of a sequence is $u _ { n }$.\\
The sequence is defined by
$$u _ { n + 1 } = p u _ { n } + q$$
where $p$ and $q$ are constants.\\
The first three terms of the sequence are given by
$$u _ { 1 } = - 8 \quad u _ { 2 } = 8 \quad u _ { 3 } = 4$$
\begin{enumerate}[label=(\alph*)]
\item Show that $q = 6$ and find the value of $p$.
\item Find the value of $u _ { 4 }$.
\item The limit of $u _ { n }$ as $n$ tends to infinity is $L$.
\begin{enumerate}[label=(\roman*)]
\item Write down an equation for $L$.
\item Hence find the value of $L$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q6 [9]}}