AS June 2024 Q4
4. Peter sets up a savings plan. He makes an initial deposit of £\(D\) and then pays in £\(M\) 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.
Given that the value of the savings plan after 1 year is £6925
| Scheme | Marks | AO |
|---|---|---|
| \(M = 150\) | B1 | 3.4 |
| (1) |
Notes
B1: CAO
| Scheme | Marks | AO |
|---|---|---|
| (aux equation \(m - 1.025 = 0 \Rightarrow\)) complementary function is \(A(1.025)^n\) | B1 | 1.1b |
| Consider a trial solution of the form \(u_n = \lambda\) so \(\lambda - 1.025\lambda = 1800 \Rightarrow \lambda = \ldots\) | M1 | 1.1b |
| General solution is \(u_n = A(1.025)^n - 72000\) | A1 | 1.1b |
| \(n = 1,\ u_1 = 6925 \Rightarrow A = \ldots\) | M1 | 3.4 |
| \(u_n = 77000(1.025)^n - 72000\) | A1 | 1.1b |
| (5) |
Notes
B1: CAO
M1: substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\)
A1: CAO
M1: using the conditions in the model to calculate \(A\)
A1: CAO (must be \(u_n =\) not \(u_{n+1} =\))
Alternative
General solution is \(u_n = A(1.025)^{n-1} - 72000\)
\(u_n = 78925(1.025)^{n-1} - 72000\)
| Scheme | Marks | AO |
|---|---|---|
| \(D = 77000 - 72000 \Rightarrow D = \)£\(5000\) | B1ft | 1.1b |
| (1) |
Notes
B1ft: substitutes \(n = 0\) in their solution to find \(D\)
Alternative
\(6925 = 1.025D + 1800 \quad \Rightarrow D = 5000\)
| Scheme | Marks | AO |
|---|---|---|
| \(77000(1.025)^n - 72000 \gt 20000\) | M1 | 1.1b |
| \((1.025)^n \gt \frac{92}{77} \Rightarrow n\log(1.025) \gt \log\left(\frac{92}{77}\right)\) | M1 | 1.1b |
| \(n \gt 7.20795\ldots \Rightarrow n = 8\) | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
M1: sets their particular solution greater than 20000– their particular solution must be of the correct form \(\left(u_n = c(1.025)^n \pm d \text{ or } u_n = c(1.025)^{n-1} \pm d\right)\)
M1: dependent on previous M mark – correctly re-arranges and then applies the process of taking logs for their particular solution
A1: CAO
Alternative
\(78925(1.025)^{n-1} - 72000 \gt 20000\)
\((1.025)^{n-1} \gt \dfrac{3680}{3157} \Rightarrow (n-1)\log(1.025) \gt \log\left(\dfrac{3680}{3157}\right)\)
\(n - 1 \gt 6.20795\ldots \Rightarrow n = 8\)