Question 4 - A-Level Maths

Edexcel FP2 AS 2023 June — Question 4 9 marks

Exam Board	Edexcel
Module	FP2 AS (Further Pure 2 AS)
Year	2023
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and series, recurrence and convergence
Type	Recurrence relation solving for closed form
Difficulty	Standard +0.3 This is a straightforward application of first-order linear recurrence relations with a real-world context. Part (a) requires simple interpretation of the model (A=50, 1.005 represents 0.5% interest), part (b) is a standard recurrence relation solution using the complementary function plus particular integral method taught in FP2, and part (c) involves solving U_n=0. While it requires multiple steps, each is routine for FP2 students with no novel insight needed—slightly easier than average due to the guided structure and familiar loan context.
Spec	1.04e Sequences: nth term and recurrence relations

A student takes out a loan for $\pounds 1000$ from a bank.

The bank charges $0.5 \%$ monthly interest on the amount of the loan yet to be repaid.
At the end of each month

the interest is added to the loan
the student then repays $\pounds 50$

Let $U _ { n }$ be the amount of money owed $n$ months after the loan was taken out.
The amount of money owed by the student is modelled by the recurrence relation $$U _ { n } = 1.005 U _ { n - 1 } - A \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$ where $A$ is a constant.

1. State the value of the constant $A$.
2. Explain, in the context of the problem, the value 1.005 Using the value of $A$ found in part (a)(i),
solve the recurrence relation $$U _ { n } = 1.005 U _ { n - 1 } - A \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$
Hence determine, according to the model, the number of months it will take to completely repay the loan.

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$A = 50$	B1	3.3 - Uses the model to state $A = 50$
Interest rate is $0.5\%$ so multiplied by $1.005$	B1	2.4
(2 marks)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda$	M1	3.1a - A complete method to solve the recurrence relation
$\text{PS} = l \Rightarrow l = 1.005l - \text{"50"}$ leading to $l = \ldots$	M1	1.1b
$l = 10000$	A1	1.1b
$1000 = c(1.005)^0 + \text{"10000"}$ leading to $c = \ldots\{-9000\}$	M1	1.1b - Uses $U_0 = 1000$ and their $l$ to find $c$
$U_n = 10000 - 9000(1.005)^n$	A1	1.1b
(5 marks)

Alternative:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$U_n = c(1.005)^n + \lambda$	M1	3.1a
$U_0 = c(1.005)^0 + \lambda \Rightarrow 1000 = c + \lambda$	M1	1.1b
$U_1 = 1.005(1000) - 50 = 955$	M1	1.1b
$U_1 = c(1.005)^1 + \lambda \Rightarrow 955 = 1.005c + \lambda$	A1	1.1b
$U_n = 10000 - 9000(1.005)^n$	A1	1.1b
(5 marks)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$10000 - 9000(1.005)^n = 0 \Rightarrow (1.005)^n = \frac{10000}{9000} \Rightarrow n = \frac{\log\frac{10}{9}}{\log 1.005}$ or $n = \log_{1.005}\frac{10}{9}$	M1	3.4
$n = 21.1$ therefore $22$	A1	3.2a
(2 marks)

Question (b) — Recurrence Relation

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Complete method to solve using $U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda$	M1
Uses value of $A$ with $PS = l$, i.e. $l = 1.005l - 50$ to find $l$	M1
$l = 10000$	A1
Uses $U_0 = 1000$ and their $l$ to find $c$	M1
$U_n = 10000 - 9000(1.005)^n$	A1	Fully correctly defined sequence

Alternative:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Uses $U_0$ to form equation for $c$ and $\lambda$	M1
Uses given formula to find $U_1$, then forms another equation for $c$ and $\lambda$	M1
Correct second equation using $U_1$	A1
Solve simultaneously to find $U_n = 10000 - 9000(1.005)^n$	A1

Question (c) — Finding Number of Months

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Sets expression $= 0$ and solves using logarithms to find $n$	M1
$n = 22$ (months)	A1	Using trial and error must show $U_{21} > 0$ and $U_{22} < 0$

## Question 4:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = 50$ | B1 | 3.3 - Uses the model to state $A = 50$ |
| Interest rate is $0.5\%$ so multiplied by $1.005$ | B1 | 2.4 |
| **(2 marks)** | | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda$ | M1 | 3.1a - A complete method to solve the recurrence relation |
| $\text{PS} = l \Rightarrow l = 1.005l - \text{"50"}$ leading to $l = \ldots$ | M1 | 1.1b |
| $l = 10000$ | A1 | 1.1b |
| $1000 = c(1.005)^0 + \text{"10000"}$ leading to $c = \ldots\{-9000\}$ | M1 | 1.1b - Uses $U_0 = 1000$ and their $l$ to find $c$ |
| $U_n = 10000 - 9000(1.005)^n$ | A1 | 1.1b |
| **(5 marks)** | | |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_n = c(1.005)^n + \lambda$ | M1 | 3.1a |
| $U_0 = c(1.005)^0 + \lambda \Rightarrow 1000 = c + \lambda$ | M1 | 1.1b |
| $U_1 = 1.005(1000) - 50 = 955$ | M1 | 1.1b |
| $U_1 = c(1.005)^1 + \lambda \Rightarrow 955 = 1.005c + \lambda$ | A1 | 1.1b |
| $U_n = 10000 - 9000(1.005)^n$ | A1 | 1.1b |
| **(5 marks)** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $10000 - 9000(1.005)^n = 0 \Rightarrow (1.005)^n = \frac{10000}{9000} \Rightarrow n = \frac{\log\frac{10}{9}}{\log 1.005}$ or $n = \log_{1.005}\frac{10}{9}$ | M1 | 3.4 |
| $n = 21.1$ therefore $22$ | A1 | 3.2a |
| **(2 marks)** | | |

# Question (b) — Recurrence Relation

| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to solve using $U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda$ | M1 | |
| Uses value of $A$ with $PS = l$, i.e. $l = 1.005l - 50$ to find $l$ | M1 | |
| $l = 10000$ | A1 | |
| Uses $U_0 = 1000$ and their $l$ to find $c$ | M1 | |
| $U_n = 10000 - 9000(1.005)^n$ | A1 | Fully correctly defined sequence |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $U_0$ to form equation for $c$ and $\lambda$ | M1 | |
| Uses given formula to find $U_1$, then forms another equation for $c$ and $\lambda$ | M1 | |
| Correct second equation using $U_1$ | A1 | |
| Solve simultaneously to find $U_n = 10000 - 9000(1.005)^n$ | A1 | |

---

# Question (c) — Finding Number of Months

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets expression $= 0$ and solves using logarithms to find $n$ | M1 | |
| $n = 22$ (months) | A1 | Using trial and error must show $U_{21} > 0$ and $U_{22} < 0$ |

---

Show LaTeX source

\begin{enumerate}
  \item A student takes out a loan for $\pounds 1000$ from a bank.
\end{enumerate}

The bank charges $0.5 \%$ monthly interest on the amount of the loan yet to be repaid.\\
At the end of each month

\begin{itemize}
  \item the interest is added to the loan
  \item the student then repays $\pounds 50$
\end{itemize}

Let $U _ { n }$ be the amount of money owed $n$ months after the loan was taken out.\\
The amount of money owed by the student is modelled by the recurrence relation

$$U _ { n } = 1.005 U _ { n - 1 } - A \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$

where $A$ is a constant.\\
(a) (i) State the value of the constant $A$.\\
(ii) Explain, in the context of the problem, the value 1.005

Using the value of $A$ found in part (a)(i),\\
(b) solve the recurrence relation

$$U _ { n } = 1.005 U _ { n - 1 } - A \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$

(c) Hence determine, according to the model, the number of months it will take to completely repay the loan.

\hfill \mbox{\textit{Edexcel FP2 AS 2023 Q4 [9]}}

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