| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compare two growth models |
| Difficulty | Moderate -0.3 This is a straightforward application of arithmetic and geometric sequences with clear scaffolding. Part (a) requires finding the common difference of an AP, part (b) finding the common ratio of a GP, and part (c) applying standard sum formulas. While it involves multiple steps and two models, each part uses direct formula application with no conceptual challenges or novel problem-solving required—slightly easier than a typical A-level question due to its routine nature. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Attempts to use \(31\,500 = 16\,200 + 9d\) to find 'd' | M1 | Attempts to use the AP formula in an attempt to find 'd'. Accept an attempt at \(31\,500 = 16\,200 + 9d\) resulting in a value for \(d\). |
| For \(16\,200 +\) their \(d = (1700)\) where \(d\) has been found by an allowable method | M1 | A correct attempt to find the second term by adding \(16\,200\) to their 'd' which must have been found via an allowable method. Allow \(d\) to be found from an "incorrect" AP formula with \(10d\) being used instead of \(9d\). Eg \(31\,500 = 16\,200 + 10d\) or more likely \(\frac{31\,500 - 16\,200}{9} = 1530\) usually leading to an answer of \(17\,730\) |
| Year 2 salary is \((£)17\,900\) | A1 | Year 2 salary is \((£) 17\,900\) |
| (3 marks) | ||
| (b) Attempts to use \(31\,500 = 16\,200^r\) to find 'r' | M1 | Attempts to use the GP formula in an attempt to find 'r'. Accept an attempt at \(31\,500 = 16\,200^r \Rightarrow r^9 = \frac{31\,500}{16\,200} \Rightarrow r = ...\) condoning numerical slips. |
| For \(16\,200 \times\) their \(r = (1.077)\) where \(r\) has been found by an allowable method | M1 | A correct attempt to find the second term by multiplying \(16\,200\) by their 'r' which must have been found via an allowable method. Allow \(r\) to be found from an "incorrect" GP formula with 10 being used instead of 9. Eg following \(31\,500 = 16\,200 r^{10}\) or \(\sqrt[10]{\frac{31\,500}{16\,200}}\). You may also award, condoning slips, for an attempt at \(16\,200 \times r\) where \(r\) is their solution of \(31\,500 = 16\,200r^n\) where \(n = 9\) or 10 |
| Year 2 salary in the range \(17\,440 \leq \\)\leq 17\,450\( | A1 | For an answer in the range \)£17\,440 \leq \\(\leq 17\,450\). Note that \(r = 1.077 \Rightarrow 17\,447.40\) |
| (3 marks) | ||
| (c) Attempts \(\frac{10}{2}\{16\,200 + 31\,500\}\) or \(\frac{16\,200(1.077^{10} - 1)}{1.077 - 1}\) | M1 |
**(a)** Attempts to use $31\,500 = 16\,200 + 9d$ to find 'd' | M1 | Attempts to use the AP formula in an attempt to find 'd'. Accept an attempt at $31\,500 = 16\,200 + 9d$ resulting in a value for $d$.
For $16\,200 +$ their $d = (1700)$ where $d$ has been found by an allowable method | M1 | A correct attempt to find the second term by adding $16\,200$ to their 'd' which must have been found via an allowable method. Allow $d$ to be found from an "incorrect" AP formula with $10d$ being used instead of $9d$. Eg $31\,500 = 16\,200 + 10d$ or more likely $\frac{31\,500 - 16\,200}{9} = 1530$ usually leading to an answer of $17\,730$
Year 2 salary is $(£)17\,900$ | A1 | Year 2 salary is $(£) 17\,900$
| (3 marks)
**(b)** Attempts to use $31\,500 = 16\,200^r$ to find 'r' | M1 | Attempts to use the GP formula in an attempt to find 'r'. Accept an attempt at $31\,500 = 16\,200^r \Rightarrow r^9 = \frac{31\,500}{16\,200} \Rightarrow r = ...$ condoning numerical slips.
For $16\,200 \times$ their $r = (1.077)$ where $r$ has been found by an allowable method | M1 | A correct attempt to find the second term by multiplying $16\,200$ by their 'r' which must have been found via an allowable method. Allow $r$ to be found from an "incorrect" GP formula with 10 being used instead of 9. Eg following $31\,500 = 16\,200 r^{10}$ or $\sqrt[10]{\frac{31\,500}{16\,200}}$. You may also award, condoning slips, for an attempt at $16\,200 \times r$ where $r$ is their solution of $31\,500 = 16\,200r^n$ where $n = 9$ or 10
Year 2 salary in the range $17\,440 \leq \$\leq 17\,450$ | A1 | For an answer in the range $£17\,440 \leq \$\leq 17\,450$. Note that $r = 1.077 \Rightarrow 17\,447.40$
| (3 marks)
**(c)** Attempts $\frac{10}{2}\{16\,200 + 31\,500\}$ or $\frac{16\,200(1.077^{10} - 1)}{1.077 - 1}$ | M1 |7. Kim starts working for a company.
\begin{itemize}
\item In year 1 her annual salary will be $\pounds 16200$
\item In year 10 her annual salary is predicted to be $\pounds 31500$
\end{itemize}
Model $A$ assumes that her annual salary will increase by the same amount each year.
\begin{enumerate}[label=(\alph*)]
\item According to model $A$, determine Kim's annual salary in year 2 .
Model $B$ assumes that her annual salary will increase by the same percentage each year.
\item According to model $B$, determine Kim's annual salary in year 2 . Give your answer to the nearest $\pounds 10$
\item Calculate, according to the two models, the difference between the total amounts that Kim is predicted to earn from year 1 to year 10 inclusive. Give your answer to the nearest £10
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q7 [9]}}