| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find term or total |
| Difficulty | Moderate -0.8 This is a straightforward application of arithmetic and geometric sequences with clear setup. Part (a) requires finding the common difference and using the nth term formula for an AP. Part (b) requires finding the common ratio and using the nth term formula for a GP. Both are standard textbook exercises with no conceptual challenges or novel problem-solving required. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Ben's profits can be modelled by an arithmetic progression, therefore has \(n^{th}\) term \(a + (n-1)d\). \(a = 24000\), \(d = \frac{64000 - 24000}{10} = 4000\) | M1 | Using the \(n^{th}\) term \(= a + (n-1)d\) of an A.P. and attempts to find value of \(d\) |
| \(5^{th} \text{ term}: 24000 + (5-1)4000 = 40000\) | M1 | Uses \(a + 4d\) with \(a = 24000\) and \(d = \ldots (4000)\) to find the profit in Year 5 |
| A1 | £40000 |
| Answer | Marks | Guidance |
|---|---|---|
| In this case, Ben's profits can be modelled with a geometric progression, with \(n^{th}\) term \(ar^{n-1}\) | M1 | Using the \(n^{th}\) term \(= ar^{n-1}\) of a G.P. and attempts to find \(r\) |
| \(a = 24000 \Rightarrow 24000r^{10} = 64000 \Rightarrow r = \sqrt[10]{\frac{64000}{24000}} = \left(\frac{64}{24}\right)^{\frac{1}{10}}\) | ||
| In year 5, profit \(= 24000\left(\left(\frac{64}{24}\right)^{\frac{1}{10}}\right)^4 = £35530.29 = £35530\) to the nearest £10 | M1 | Uses \(ar^4\) with \(a = 24000\) and \(r = \left(\frac{64}{24}\right)^{\frac{1}{10}}\) to find the profit in year 5 |
| A1 | £35530 |
**Part a:** According to model P, determine Ben's profit in year 5
| Ben's profits can be modelled by an arithmetic progression, therefore has $n^{th}$ term $a + (n-1)d$. $a = 24000$, $d = \frac{64000 - 24000}{10} = 4000$ | M1 | Using the $n^{th}$ term $= a + (n-1)d$ of an A.P. and attempts to find value of $d$ |
| --- | --- | --- |
| $5^{th} \text{ term}: 24000 + (5-1)4000 = 40000$ | M1 | Uses $a + 4d$ with $a = 24000$ and $d = \ldots (4000)$ to find the profit in Year 5 |
| | A1 | £40000 |
**Part b:** According to model Q, determine Ben's profit in year 5. Give your answer to the nearest £10
| In this case, Ben's profits can be modelled with a geometric progression, with $n^{th}$ term $ar^{n-1}$ | M1 | Using the $n^{th}$ term $= ar^{n-1}$ of a G.P. and attempts to find $r$ |
| --- | --- | --- |
| $a = 24000 \Rightarrow 24000r^{10} = 64000 \Rightarrow r = \sqrt[10]{\frac{64000}{24000}} = \left(\frac{64}{24}\right)^{\frac{1}{10}}$ | | |
| In year 5, profit $= 24000\left(\left(\frac{64}{24}\right)^{\frac{1}{10}}\right)^4 = £35530.29 = £35530$ to the nearest £10 | M1 | Uses $ar^4$ with $a = 24000$ and $r = \left(\frac{64}{24}\right)^{\frac{1}{10}}$ to find the profit in year 5 |
| | A1 | £35530 |
**(3) marks + (3) marks = (6) marks total for Question 4**
---\begin{enumerate}
\item Ben starts a new company.
\end{enumerate}
\begin{itemize}
\item In year 1 his profits will be $\pounds 24000$.
\item In year 11 his profit is predicted to be $\pounds 64000$.
\end{itemize}
Model $\boldsymbol { P }$ assumes that his profit will increase by the same amount each year.\\
a. According to model $\boldsymbol { P }$, determine Ben's profit in year 5.
Model $\boldsymbol { Q }$ assumes that his profit will increase by the same percentage each year.\\
b. According to model $\boldsymbol { Q }$, determine Ben's profit in year 5 . Give your answer to the nearest £10.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q4 [6]}}