| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: find specific terms |
| Difficulty | Moderate -0.5 This is a straightforward recurrence relation question requiring simple iterative calculations and basic algebraic manipulation. Part (a) involves three direct substitutions, part (b) requires continuing the iteration until the population becomes negative, and part (c) solves a simple equation by setting u_{n+1} = u_n. While it involves multiple parts, each step is routine with no conceptual challenges beyond understanding what a recurrence relation represents—slightly easier than average for A-level. |
| Spec | 1.02z Models in context: use functions in modelling1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_1 = 1.05(500000) - 15000 = 510000\) | B1 | |
| \(u_2 = 1.05(510000) - 15000 = 520500\) | B1 | |
| \(u_3 = 1.05(520500) - 15000 = 531525\) | B1 | |
| Population is increasing | B1 | Comment required |
| Answer | Marks |
|---|---|
| \(u_1 = 1.05(500000)-100000 = 425000\) | M1 |
| \(u_2 = 346250,\ u_3 = 263562.5,\ u_4 = 176740.6,\ u_5 = 85577.6\) | A1 |
| \(u_6 = 1.05(85577.6)-100000 < 0\) (dies out in sixth year) | A1 |
| Answer | Marks |
|---|---|
| For unchanged population: \(u_{n+1} = u_n\), so \(1.05u_n - d = u_n\) | M1 |
| \(0.05u_n = d \Rightarrow d = 0.05 \times 500000 = 25000\) | A1 |
# Question 7:
## Part (a):
| $u_1 = 1.05(500000) - 15000 = 510000$ | B1 | |
| $u_2 = 1.05(510000) - 15000 = 520500$ | B1 | |
| $u_3 = 1.05(520500) - 15000 = 531525$ | B1 | |
| Population is increasing | B1 | Comment required |
## Part (b):
| $u_1 = 1.05(500000)-100000 = 425000$ | M1 | |
| $u_2 = 346250,\ u_3 = 263562.5,\ u_4 = 176740.6,\ u_5 = 85577.6$ | A1 | |
| $u_6 = 1.05(85577.6)-100000 < 0$ (dies out in sixth year) | A1 | |
## Part (c):
| For unchanged population: $u_{n+1} = u_n$, so $1.05u_n - d = u_n$ | M1 | |
| $0.05u_n = d \Rightarrow d = 0.05 \times 500000 = 25000$ | A1 | |
---7. Initially the number of fish in a lake is 500000 . The population is then modelled by the recurrence relation
$$u _ { n + 1 } = 1.05 u _ { n } - d , \quad u _ { 0 } = 500000 .$$
In this relation $u _ { n }$ is the number of fish in the lake after $n$ years and $d$ is the number of fish which are caught each year.
Given that $d = 15000$,
\begin{enumerate}[label=(\alph*)]
\item calculate $u _ { 1 } , u _ { 2 }$ and $u _ { 3 }$ and comment briefly on your results.
Given that $d = 100000$,
\item show that the population of fish dies out during the sixth year.
\item Find the value of $d$ which would leave the population each year unchanged.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q7 [8]}}