| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Total over time period |
| Difficulty | Standard +0.3 This is a straightforward geometric sequence question with standard applications: verifying a term, identifying the common ratio, manipulating logarithms in an inequality, and finding a sum. Part (c) is scaffolded and part (e) requires the GP sum formula, but all steps are routine C2 techniques with no novel insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(25000 \times 1.03 = 25750\); \(\left\{25000+750=25750, \text{ or } 25000\frac{(1-0.03^2)}{1-0.03}=25750\right\}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r=1.03\); allow \(\frac{103}{100}\) or \(1\frac{3}{100}\) but no other alternatives | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(25000r^{N-1}>40000\) (either letter \(r\) or their \(r\) value); allow \('='\) or \('<'\) | M1 | |
| \(r^M>1.6 \Rightarrow \log r^M > \log 1.6\); allow \('='\) or \('<'\) | M1 | |
OR by change of base: \(\log_{1.03}1.6 < M \Rightarrow \frac{\log 1.6}{\log 1.03}| M1 |
| |
| \((N-1)\log 1.03 > \log 1.6\) (correct bracketing required) | A1 cso | Accept work for part (c) seen in part (d) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt to evaluate \(\frac{\log 1.6}{\log 1.03}+1\) or \(\{25000(1.03)^{15}\) and \(25000(1.03)^{16}\}\) | M1 | |
| \(N=17\) (not 16.9 and not e.g. \(N\geq17\)); allow '17th year' | A1 | Accept work for part (d) seen in part (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using formula \(\frac{a(1-r^n)}{1-r}\) with values of \(a\) and \(r\), and \(n=9\), 10 or 11 | M1 | |
| \(\frac{25000(1-1.03^{10})}{1-1.03}\) | A1 | |
| \(287\,000\) (must be rounded to nearest 1000); allow 287000.00 | A1 |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $25000 \times 1.03 = 25750$; $\left\{25000+750=25750, \text{ or } 25000\frac{(1-0.03^2)}{1-0.03}=25750\right\}$ | B1 | |
**(1 mark)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r=1.03$; allow $\frac{103}{100}$ or $1\frac{3}{100}$ but no other alternatives | B1 | |
**(1 mark)**
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $25000r^{N-1}>40000$ (either letter $r$ or their $r$ value); allow $'='$ or $'<'$ | M1 | |
| $r^M>1.6 \Rightarrow \log r^M > \log 1.6$; allow $'='$ or $'<'$ | M1 | |
| OR by change of base: $\log_{1.03}1.6 < M \Rightarrow \frac{\log 1.6}{\log 1.03}<M$ | M1 | |
| $(N-1)\log 1.03 > \log 1.6$ (correct bracketing required) | A1 cso | Accept work for part (c) seen in part (d) |
**(3 marks)**
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to evaluate $\frac{\log 1.6}{\log 1.03}+1$ or $\{25000(1.03)^{15}$ and $25000(1.03)^{16}\}$ | M1 | |
| $N=17$ (not 16.9 and not e.g. $N\geq17$); allow '17th year' | A1 | Accept work for part (d) seen in part (c) |
**(2 marks)**
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using formula $\frac{a(1-r^n)}{1-r}$ with values of $a$ and $r$, and $n=9$, 10 or 11 | M1 | |
| $\frac{25000(1-1.03^{10})}{1-1.03}$ | A1 | |
| $287\,000$ (must be rounded to nearest 1000); allow 287000.00 | A1 | |
**(3 marks, 10 total)**
---9. The adult population of a town is 25000 at the end of Year 1.
A model predicts that the adult population of the town will increase by $3 \%$ each year, forming a geometric sequence.
\begin{enumerate}[label=(\alph*)]
\item Show that the predicted adult population at the end of Year 2 is 25750.
\item Write down the common ratio of the geometric sequence.
The model predicts that Year $N$ will be the first year in which the adult population of the town exceeds 40000.
\item Show that
$$( N - 1 ) \log 1.03 > \log 1.6$$
\item Find the value of $N$.
At the end of each year, each member of the adult population of the town will give $\pounds 1$ to a charity fund.
Assuming the population model,
\item find the total amount that will be given to the charity fund for the 10 years from the end of Year 1 to the end of Year 10, giving your answer to the nearest $\pounds 1000$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q9 [10]}}