AS June 2023 Q4
4. A student takes out a loan for £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 £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
\[U_n = 1.005U_{n-1} - A \qquad U_0 = 1000 \qquad n \in \mathbb{Z}^+\]where \(A\) is a constant.
Using the value of \(A\) found in part (a)(i),
| Scheme | Marks | AO |
|---|---|---|
| \(A = 50\) | B1 | 3.3 |
| Interest rate is 0.5% so multiplied by 1.005 | B1 | 2.4 |
| (2) |
Notes
B1: Uses the model to state \(A = 50\)
B1: A correct explanation
| Scheme | Marks | AO |
|---|---|---|
| A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\) | M1 | 3.1a |
| \(\text{PS} = \lambda \Rightarrow \lambda = 1.005\lambda - \text{“}50\text{”}\) leading to \(\lambda = \ldots\) | M1 | 1.1b |
| \(\lambda = 10000\) | A1 | 1.1b |
| Uses \(U_0 = 1000\) and their value of \(\lambda\) to find the value of \(c\) \(1000 = c(1.005)^0 + \text{“}10000\text{”}\) Leading to \(c = \ldots\{-9000\}\) | M1 | 1.1b |
| \(U_n = 10000 - 9000(1.005)^n\) | A1 | 1.1b |
| (5) |
Notes
M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\)
M1: Uses their value of \(A\) with \(\text{PS} = \lambda \Rightarrow \lambda = 1.005\lambda - \text{“}50\text{”}\) to find a value for \(\lambda\)
A1: \(\lambda = 10000\)
M1: Uses \(U_0 = 1000\) and their value of \(\lambda\) to find the value of \(c\)
A1: Fully correctly defined sequence \(U_n = 10000 - 9000(1.005)^n\)
Alternative
| Scheme | Marks | AO |
|---|---|---|
| A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\) | M1 | 3.1a |
| \(U_0 = c(1.005)^0 + \lambda \Rightarrow 1000 = c + \lambda\) | M1 | 1.1b |
| \(U_1 = 1.005(1000) - 50 = 955\) \(U_1 = c(1.005)^1 + \lambda \Rightarrow 955 = 1.005c + \lambda\) | M1 A1 | 1.1b 1.1b |
| \(U_n = 10000 - 9000(1.005)^n\) | A1 | 1.1b |
| (5) |
M1: A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.005)^n + \lambda\)
M1: Uses \(U_0\) to form an equation for their \(c\) and their \(\lambda\)
M1: Uses the given formula to find \(U_1\) and then uses their value to form another equation for their \(c\) and their \(\lambda\)
A1: Correct second equation using \(U_1\)
A1: Solve simultaneously to find \(U_n = 10000 - 9000(1.005)^n\)
| Scheme | Marks | AO |
|---|---|---|
| \(10000 - 9000(1.005)^n = 0 \Rightarrow n = \ldots\) \((1.005)^n = \dfrac{10000}{9000} \Rightarrow n = \dfrac{\log\frac{10}{9}}{\log 1.005} = \ldots\) \((1.005)^n = \dfrac{10000}{9000} \Rightarrow n = \log_{1.005}\dfrac{10}{9} = \ldots\) | M1 | 3.4 |
| \(n = 21.1\) therefore 22 | A1 | 3.2a |
| (2) | ||
| (9 marks) |
Notes
M1: Sets their expression = 0 and solves using logarithms to find a value for \(n\).
A1: 22 (months)
Note: Using trail and error must show that \(U_{21} \gt 0\) is positive and \(U_{22} \lt 0\)