| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Total over time period |
| Difficulty | Moderate -0.3 This is a straightforward geometric sequence application requiring (a) finding the 13th term using the formula with r=1.02, and (b) summing the geometric series multiplied by £5. Both parts use standard formulas with no conceptual challenges, though the multi-step calculation and context interpretation make it slightly more involved than pure recall. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(r = 1.02\) | B1 | States or uses \(r=1.02\), \(102\%\), \((1+2\%)\) oe |
| Attempts \(25000 \times 1.02^{13} = 32340\) or \(32341\) or \(32300\) | M1 A1 | Allow \(r=2\) for M1. May show increase each year. Allow misread if \(2500\) or \(250000\) used. Awrt \(31706\) (i.e. \(25000\times1.02^{12}\)) usually sufficient for M1. Must be integer; allow \(32300\) (3sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{a(r^n-1)}{r-1} = \frac{25000(1.02^{14}-1)}{1.02-1}\) or \(\frac{125000(1.02^{14}-1)}{1.02-1}\) | M1 A1 | With \(a=25000\) or \(125000\), \(n=13\) or \(14\), \(r=1.02\) or their \(r\) from (a). Awrt \(1\,997\,000\) implies both marks |
| \(£1\,997\,000\) | A1 | \(1\,997\,000\) only |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $r = 1.02$ | B1 | States or uses $r=1.02$, $102\%$, $(1+2\%)$ oe |
| Attempts $25000 \times 1.02^{13} = 32340$ or $32341$ or $32300$ | M1 A1 | Allow $r=2$ for M1. May show increase each year. Allow misread if $2500$ or $250000$ used. Awrt $31706$ (i.e. $25000\times1.02^{12}$) usually sufficient for M1. Must be integer; allow $32300$ (3sf) |
**(3 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{a(r^n-1)}{r-1} = \frac{25000(1.02^{14}-1)}{1.02-1}$ or $\frac{125000(1.02^{14}-1)}{1.02-1}$ | M1 A1 | With $a=25000$ or $125000$, $n=13$ or $14$, $r=1.02$ or their $r$ from (a). Awrt $1\,997\,000$ implies both marks |
| $£1\,997\,000$ | A1 | $1\,997\,000$ only |
**(3 marks)**2. The adult population of a town at the start of 2019 is 25000
A model predicts that the adult population will increase by $2 \%$ each year, so that the number of adults in the population at the start of each year following 2019 will form a geometric sequence.
\begin{enumerate}[label=(\alph*)]
\item Find, according to the model, the adult population of the town at the start of 2032
It is also modelled that every member of the adult population gives $\pounds 5$ to local charity at the start of each year.
\item Find, according to these models, the total amount of money that would be given to local charity by the adult population of the town from 2019 to 2032 inclusive. Give your answer to the nearest $\pounds 1000$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q2 [6]}}