AQA C2 2011 June — Question 7 6 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2011
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeLinear iterative formula u(n+1) = pu(n) + q
DifficultyStandard +0.3 This is a straightforward recurrence relation problem requiring substitution of given values into standard formulas. Part (a) uses the limit condition L = pL + q and the first two terms to form simultaneous equations—routine algebraic manipulation. Part (b) is direct substitution. While it involves multiple steps, each is mechanical with no conceptual challenges beyond knowing the limit formula for linear recurrence relations.
Spec1.04e Sequences: nth term and recurrence relations

7 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 two terms of the sequence are given by \(u _ { 1 } = 60\) and \(u _ { 2 } = 48\).
The limit of \(u _ { n }\) as \(n\) tends to infinity is 12 .
  1. Show that \(p = \frac { 3 } { 4 }\) and find the value of \(q\).
  2. Find the value of \(u _ { 3 }\).

Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(48 = 60p + q\)M1 M1 for each equation in ACF (Condone embedded values for M1M1)
\(12 = 12p + q\)M1
\(36 = 48p\) or \(p = \frac{36}{48}\)m1 Valid method to solve the correct two simultaneous equations in \(p\) and \(q\) to at least the stage \(ap = b\) or \(cq = d\)
\(p = \frac{3}{4}\)A1 AG (condone if left as equivalent decimal)
\(q = 3\)B1 (Total: 5) Can award if seen explicitly in (b) and no contradiction [i.e. not attempted in (a)]
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(u_3 = 36 + q = 39\)B1F (Total: 1) If not 39, ft on \((36 +\) c's \(q)\)
## Question 7:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $48 = 60p + q$ | M1 | M1 for each equation in ACF (Condone embedded values for M1M1) |
| $12 = 12p + q$ | M1 | |
| $36 = 48p$ or $p = \frac{36}{48}$ | m1 | Valid method to solve the correct two simultaneous equations in $p$ and $q$ to at least the stage $ap = b$ or $cq = d$ |
| $p = \frac{3}{4}$ | A1 | AG (condone if left as equivalent decimal) |
| $q = 3$ | B1 (Total: 5) | Can award if seen explicitly in (b) and no contradiction [i.e. not attempted in (a)] |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_3 = 36 + q = 39$ | B1F (Total: 1) | If not 39, ft on $(36 +$ c's $q)$ |

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7 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 two terms of the sequence are given by $u _ { 1 } = 60$ and $u _ { 2 } = 48$.\\
The limit of $u _ { n }$ as $n$ tends to infinity is 12 .
\begin{enumerate}[label=(\alph*)]
\item Show that $p = \frac { 3 } { 4 }$ and find the value of $q$.
\item Find the value of $u _ { 3 }$.
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2011 Q7 [6]}}