Question 11 - A-Level Maths

AQA Paper 1 2023 June — Question 11 9 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2023
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	Standard +0.3 This is a straightforward recurrence relation question requiring routine substitution, solving a quadratic equation, and understanding convergence to a limit. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average.
Spec	1.04e Sequences: nth term and recurrence relations 1.04f Sequence types: increasing, decreasing, periodic

11 The $n$th term of a sequence is $u _ { n }$ The sequence is defined by $$u _ { n + 1 } = p u _ { n } + 70$$ where $u _ { 1 } = 400$ and $p$ is a constant.
11

Find an expression, in terms of $p$, for $u _ { 2 }$ 11
It is given that $u _ { 3 } = 382$ 11 (b) (i) Show that $p$ satisfies the equation $$200 p ^ { 2 } + 35 p - 156 = 0$$ 11 (b) (ii) It is given that the sequence is a decreasing sequence. Find the value of $u _ { 4 }$ and the value of $u _ { 5 }$ 11
The limit of $u _ { n }$ as $n$ tends to infinity is $L$ 11 (c) (i) Write down an equation for $L$ 11 (c) (ii) Find the value of $L$

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Question 11:

11(a):

Answer	Marks	Guidance
$400p + 70$	B1	Obtains $400p + 70$

11(b)(i):

Answer	Marks	Guidance
$382 = pu_2 + 70$	M1	Substitutes 382 or their $u_2$ into $u_3 = pu_2 + 70$
$382 = p(400p+70)+70$; $400p^2 + 70p + 70$; $400p^2 + 70p - 312 = 0$; $200p^2 + 35p - 156 = 0$	M1	Substitutes 382 and their $u_2$ into $u_3 = pu_2 + 70$ to obtain quadratic in $p$; PI by $382 = p(400p+70)+70$
$200p^2 + 35p - 156 = 0$	R1	Obtains correct equation and rearranges to obtain given answer; must see brackets expanded before given answer

11(b)(ii):

Answer	Marks	Guidance
$p = 0.8,\ p = -0.975$	B1	Obtains both $p = 0.8$ and $-0.975$; PI by correct $u_4 = 375.6$ and $u_5 = 370.48$
$p = -0.975 \Rightarrow u_4 = -302.45,\ u_5 = 364.88875$ not decreasing	M1	Uses $p = 0.8$ or $-0.975$ to obtain a value for $u_4$; PI by $375.6$, $-302.45$, $370.48$; accept equivalent fractions or AWRT $364.89$
$u_4 = 375.6$ and $u_5 = 370.48$	R1	Deduces correct values for $u_4$ and $u_5$; $(u_4=)375.6$ and $(u_5=)370.48$; accept equivalent fractions; if incorrect values are seen they must be rejected

11(c)(i):

Answer	Marks	Guidance
$L = 0.8L + 70$	B1	Forms the equation $L = pL + 70$ or $(1-p)L = 70$; or with $p = 0.8$ or $-0.975$ substituted into either of these equations; accept if $1-p$ is evaluated; ISW

11(c)(ii):

Answer	Marks	Guidance
$350$	R1	Deduces the value of $L$ is $350$ or AWRT $35.4$; accept $\frac{2800}{79}$ or both

## Question 11:

### 11(a):
$400p + 70$ | B1 | Obtains $400p + 70$

### 11(b)(i):
$382 = pu_2 + 70$ | M1 | Substitutes 382 **or** their $u_2$ into $u_3 = pu_2 + 70$

$382 = p(400p+70)+70$; $400p^2 + 70p + 70$; $400p^2 + 70p - 312 = 0$; $200p^2 + 35p - 156 = 0$ | M1 | Substitutes 382 **and** their $u_2$ into $u_3 = pu_2 + 70$ to obtain quadratic in $p$; PI by $382 = p(400p+70)+70$

$200p^2 + 35p - 156 = 0$ | R1 | Obtains correct equation and rearranges to obtain given answer; must see brackets expanded before given answer

### 11(b)(ii):
$p = 0.8,\ p = -0.975$ | B1 | Obtains both $p = 0.8$ and $-0.975$; PI by correct $u_4 = 375.6$ **and** $u_5 = 370.48$

$p = -0.975 \Rightarrow u_4 = -302.45,\ u_5 = 364.88875$ not decreasing | M1 | Uses $p = 0.8$ or $-0.975$ to obtain a value for $u_4$; PI by $375.6$, $-302.45$, $370.48$; accept equivalent fractions or AWRT $364.89$

$u_4 = 375.6$ and $u_5 = 370.48$ | R1 | Deduces correct values for $u_4$ and $u_5$; $(u_4=)375.6$ and $(u_5=)370.48$; accept equivalent fractions; if incorrect values are seen they must be rejected

### 11(c)(i):
$L = 0.8L + 70$ | B1 | Forms the equation $L = pL + 70$ or $(1-p)L = 70$; or with $p = 0.8$ or $-0.975$ substituted into either of these equations; accept if $1-p$ is evaluated; ISW

### 11(c)(ii):
$350$ | R1 | Deduces the value of $L$ is $350$ or AWRT $35.4$; accept $\frac{2800}{79}$ or both

Show LaTeX source

11 The $n$th term of a sequence is $u _ { n }$\\
The sequence is defined by

$$u _ { n + 1 } = p u _ { n } + 70$$

where $u _ { 1 } = 400$ and $p$ is a constant.\\
11
\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $p$, for $u _ { 2 }$

11
\item It is given that $u _ { 3 } = 382$\\
11 (b) (i) Show that $p$ satisfies the equation

$$200 p ^ { 2 } + 35 p - 156 = 0$$

11 (b) (ii) It is given that the sequence is a decreasing sequence.

Find the value of $u _ { 4 }$ and the value of $u _ { 5 }$\\

11
\item The limit of $u _ { n }$ as $n$ tends to infinity is $L$\\
11 (c) (i) Write down an equation for $L$\\

11 (c) (ii) Find the value of $L$
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2023 Q11 [9]}}

This paper (16 questions)

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Q1 1 Q2 1 Q3 1 Q4 1 Q5 4 Q6 5 Q7 4 Q8 6 Q9 9 Q10 8 Q11 9 Q12 8 Q13 9 Q14 13 Q15 9 Q16 14

Answer	Marks	Guidance
\(382 = pu_2 + 70\)	M1	Substitutes 382 or their \(u_2\) into \(u_3 = pu_2 + 70\)
\(382 = p(400p+70)+70\); \(400p^2 + 70p + 70\); \(400p^2 + 70p - 312 = 0\); \(200p^2 + 35p - 156 = 0\)	M1	Substitutes 382 and their \(u_2\) into \(u_3 = pu_2 + 70\) to obtain quadratic in \(p\); PI by \(382 = p(400p+70)+70\)
\(200p^2 + 35p - 156 = 0\)	R1	Obtains correct equation and rearranges to obtain given answer; must see brackets expanded before given answer

Answer	Marks	Guidance
\(p = 0.8,\ p = -0.975\)	B1	Obtains both \(p = 0.8\) and \(-0.975\); PI by correct \(u_4 = 375.6\) and \(u_5 = 370.48\)
\(p = -0.975 \Rightarrow u_4 = -302.45,\ u_5 = 364.88875\) not decreasing	M1	Uses \(p = 0.8\) or \(-0.975\) to obtain a value for \(u_4\); PI by \(375.6\), \(-302.45\), \(370.48\); accept equivalent fractions or AWRT \(364.89\)
\(u_4 = 375.6\) and \(u_5 = 370.48\)	R1	Deduces correct values for \(u_4\) and \(u_5\); \((u_4=)375.6\) and \((u_5=)370.48\); accept equivalent fractions; if incorrect values are seen they must be rejected