Question 4 - A-Level Maths

Edexcel FD2 AS 2024 June — Question 4 10 marks

Exam Board	Edexcel
Module	FD2 AS (Further Decision 2 AS)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Dynamic Programming
Type	Recurrence relation solution
Difficulty	Standard +0.3 This is a standard recurrence relation problem requiring routine techniques: identifying monthly payment from annual formula, solving a first-order linear recurrence (standard method taught in D2), substituting a boundary condition, and solving a logarithmic inequality. All steps follow textbook procedures with no novel insight required, making it slightly easier than average.
Spec	1.04e Sequences: nth term and recurrence relations

4. Peter sets up a savings plan. He makes an initial deposit of $\pounds D$ and then pays in $\pounds M$ at the end of each month. The value of the savings plan, in pounds, is modelled by $$u _ { n + 1 } = 1.025 u _ { n } + 1800$$ where $n \geqslant 0$ is an integer and $u _ { n }$ is the total value of the savings plan, in pounds, after $n$ years.

Calculate the value of $M$ Given that the value of the savings plan after 1 year is $\pounds 6925$
solve the recurrence relation for $u _ { n }$
Determine the value of $D$
Hence determine, using algebra, the number of years it will take for the value of the savings plan to exceed $\pounds 20000$

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$M = 150$	B1

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Auxiliary equation $m - 1.025 = 0$; complementary function is $A(1.025)^n$	B1
Trial solution $u_n = \lambda$, so $\lambda - 1.025\lambda = 1800 \Rightarrow \lambda = \ldots$	M1
General solution is $u_n = A(1.025)^n - 72000$	A1
$n=1$, $u_1 = 6925 \Rightarrow A = \ldots$	M1
$u_n = 77000(1.025)^n - 72000$	A1

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$D = 77000 - 72000 \Rightarrow D = \pounds 5000$	B1ft	ft their answer; alternatively $6925 = 1.025D + 1800 \Rightarrow D = 5000$

Part (d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$77000(1.025)^n - 72000 > 20000$	M1
$(1.025)^n > \frac{92}{77} \Rightarrow n\log(1.025) > \log\left(\frac{92}{77}\right)$	M1
$n > 7.20795\ldots \Rightarrow n = 8$	A1

Mark Scheme

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Correct answer only	B1	CAO

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Correct answer only	B1	CAO
Substituting trial solution into the recurrence relation to find $\lambda$	M1	Attempt to find $\lambda$
Correct answer only	A1	CAO
Using the conditions in the model to calculate $A$	M1
Correct answer only	A1	CAO; must be $u_n =$ not $u_{n+1} =$

Part (c)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Substitutes $n = 0$ into their solution to find $D$	B1ft	Follow through on their solution

Part (d)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Sets their particular solution $> 20000$	M1	Particular solution must be of correct form: $u_n = c(1.025)^n \pm d$ or $u_n = c(1.025)^{n-1} \pm d$
Correctly re-arranges and applies logs to their particular solution	M1	Dependent on previous M mark
Correct answer only	A1	CAO

# Question 4:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $M = 150$ | B1 | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Auxiliary equation $m - 1.025 = 0$; complementary function is $A(1.025)^n$ | B1 | |
| Trial solution $u_n = \lambda$, so $\lambda - 1.025\lambda = 1800 \Rightarrow \lambda = \ldots$ | M1 | |
| General solution is $u_n = A(1.025)^n - 72000$ | A1 | |
| $n=1$, $u_1 = 6925 \Rightarrow A = \ldots$ | M1 | |
| $u_n = 77000(1.025)^n - 72000$ | A1 | |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $D = 77000 - 72000 \Rightarrow D = \pounds 5000$ | B1ft | ft their answer; alternatively $6925 = 1.025D + 1800 \Rightarrow D = 5000$ |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $77000(1.025)^n - 72000 > 20000$ | M1 | |
| $(1.025)^n > \frac{92}{77} \Rightarrow n\log(1.025) > \log\left(\frac{92}{77}\right)$ | M1 | |
| $n > 7.20795\ldots \Rightarrow n = 8$ | A1 | |

# Mark Scheme

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct answer only | **B1** | CAO |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct answer only | **B1** | CAO |
| Substituting trial solution into the recurrence relation to find $\lambda$ | **M1** | Attempt to find $\lambda$ |
| Correct answer only | **A1** | CAO |
| Using the conditions in the model to calculate $A$ | **M1** | |
| Correct answer only | **A1** | CAO; must be $u_n =$ not $u_{n+1} =$ |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $n = 0$ into their solution to find $D$ | **B1ft** | Follow through on their solution |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets their particular solution $> 20000$ | **M1** | Particular solution must be of correct form: $u_n = c(1.025)^n \pm d$ or $u_n = c(1.025)^{n-1} \pm d$ |
| Correctly re-arranges and applies logs to their particular solution | **M1** | Dependent on previous M mark |
| Correct answer only | **A1** | CAO |

Show LaTeX source

4. Peter sets up a savings plan. He makes an initial deposit of $\pounds D$ and then pays in $\pounds M$ at the end of each month.

The value of the savings plan, in pounds, is modelled by

$$u _ { n + 1 } = 1.025 u _ { n } + 1800$$

where $n \geqslant 0$ is an integer and $u _ { n }$ is the total value of the savings plan, in pounds, after $n$ years.
\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $M$

Given that the value of the savings plan after 1 year is $\pounds 6925$
\item solve the recurrence relation for $u _ { n }$
\item Determine the value of $D$
\item Hence determine, using algebra, the number of years it will take for the value of the savings plan to exceed $\pounds 20000$
\end{enumerate}

\hfill \mbox{\textit{Edexcel FD2 AS 2024 Q4 [10]}}

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