| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric sequences with standard compound growth formulas. Part (a) is direct substitution into GP formula, part (b) requires solving simultaneous equations using the GP formula (standard technique), and part (c) involves equating two expressions and solving logarithmically. All techniques are routine for P2 level with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2000 \times 1.03^6 = \text{awrt } 2390\) | M1, A1 | M1: Attempts \(2000\times1.03^5\) or \(2000\times1.03^6\); may use successive iterations. A1: awrt 2390 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3690 = ab^4,\ 3470 = ab^7 \Rightarrow b^3 = \frac{3470}{3690} = (0.94...)\) | M1 | Makes progress using the model to find one of \(a\) or \(b^3\) (or \(b\)). Allow if \(r\) used instead of \(b\). Allow from indexing error \(3690=ab^3, 3470=ab^6\) |
| \(b = \sqrt[3]{\frac{3470}{3690}} = 0.9797...\quad a = \frac{3690}{0.9797...^4}\) | dM1 | Full method to find both \(a\) and \(b\). Condone from \(3690=ab^3, 3470=ab^6\) |
| \(N = 4000 \times 0.98^t\) | A1 | Accept awrt 4000 to 2sf, awrt 0.98. Must be an equation not just values. Condone \(n\) or \(T\) as variable |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(2000\times1.03^T = 4000\times0.98^T \Rightarrow \left(\frac{1.03}{0.98}\right)^T = \frac{4000}{2000}\) | M1 | Makes progress forming single term with single power \(T\). Follow through on their equations of correct form. Alternatively takes logs of both sides and uses addition and power laws |
| \(T\log\left(\frac{1.03}{0.98}\right) = \log 2 \Rightarrow T = ...\) | dM1 | Full and complete attempt to find \(T\). Dependent on previous method; may be implied by correct value for their \(a=b^T\) |
| \(T = 13.9\) | A1 | cao |
## Question 10:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2000 \times 1.03^6 = \text{awrt } 2390$ | M1, A1 | M1: Attempts $2000\times1.03^5$ or $2000\times1.03^6$; may use successive iterations. A1: awrt 2390 |
**(2 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3690 = ab^4,\ 3470 = ab^7 \Rightarrow b^3 = \frac{3470}{3690} = (0.94...)$ | M1 | Makes progress using the model to find one of $a$ or $b^3$ (or $b$). Allow if $r$ used instead of $b$. Allow from indexing error $3690=ab^3, 3470=ab^6$ |
| $b = \sqrt[3]{\frac{3470}{3690}} = 0.9797...\quad a = \frac{3690}{0.9797...^4}$ | dM1 | Full method to find both $a$ and $b$. Condone from $3690=ab^3, 3470=ab^6$ |
| $N = 4000 \times 0.98^t$ | A1 | Accept awrt 4000 to 2sf, awrt 0.98. Must be an equation not just values. Condone $n$ or $T$ as variable |
**(3 marks)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $2000\times1.03^T = 4000\times0.98^T \Rightarrow \left(\frac{1.03}{0.98}\right)^T = \frac{4000}{2000}$ | M1 | Makes progress forming single term with single power $T$. Follow through on their equations of correct form. Alternatively takes logs of both sides and uses addition and power laws |
| $T\log\left(\frac{1.03}{0.98}\right) = \log 2 \Rightarrow T = ...$ | dM1 | Full and complete attempt to find $T$. Dependent on previous method; may be implied by correct value for their $a=b^T$ |
| $T = 13.9$ | A1 | cao |
**(3 marks total: 8 marks)**\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
The number of dormice and the number of voles on an island are being monitored.\\
Initially there are 2000 dormice on the island.\\
A model predicts that the number of dormice will increase by $3 \%$ each year, so that the numbers of dormice on the island at the end of each year form a geometric sequence.\\
(a) Find, according to the model, the number of dormice on the island 6 years after monitoring began. Give your answer to 3 significant figures.
The number of voles on the island is being monitored over the same period of time.\\
Given that
\begin{itemize}
\item 4 years after monitoring began there were 3690 voles on the island
\item 7 years after monitoring began there were 3470 voles on the island
\item the number of voles on the island at the end of each year is modelled as a geometric sequence\\
(b) find the equation of this model in the form
\end{itemize}
$$N = a b ^ { t }$$
where $N$ is the number of voles, $t$ years after monitoring began and $a$ and $b$ are constants. Give the value of $a$ and the value of $b$ to 2 significant figures.
When $t = T$, the number of dormice on the island is equal to the number of voles on the island.\\
(c) Find, according to the models, the value of $T$, giving your answer to one decimal place.
\hfill \mbox{\textit{Edexcel P2 2024 Q10 [8]}}