| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find year when threshold exceeded |
| Difficulty | Moderate -0.3 This is a straightforward geometric sequence application with standard bookwork. Part (a) is direct recall, part (b) is a routine logarithm manipulation shown step-by-step, parts (c) and (d) apply standard GP formulas with calculator work. The multi-part structure adds length but not conceptual difficulty—all techniques are standard C2 material with no novel problem-solving required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \(50\,000r^{n-1}\) (or equiv.) | B1 | Allow \(ar^{n-1}\) if \(50\,000r^{n-1}\) is seen in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| \(50\,000r^{n-1} > 200\,000\) | M1 | Allow equals sign or wrong inequality sign; condone slips |
| \(r^{n-1} > 4 \Rightarrow (n-1)\log r > \log 4\) | M1 | Introducing logs and dealing correctly with the power |
| \(n > \frac{\log 4}{\log r} + 1\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 1.09\): \(n > \frac{\log 4}{\log 1.09} + 1\) or \(n - 1 > \frac{\log 4}{\log 1.09}\) (\(n > 17.086...\)) | M1 | Allow equality |
| Year 18 or 2023 | A1 | If one of these is correct, ignore the other |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_n = \frac{a(1-r^n)}{1-r} = \frac{50000(1-1.09^{10})}{1-1.09}\) | M1 A1 | |
| £760,000 | A1 | Must be this answer, nearest £10000 |
## Question 8:
### Part (a):
| $50\,000r^{n-1}$ (or equiv.) | B1 | Allow $ar^{n-1}$ if $50\,000r^{n-1}$ is seen in (b) |
### Part (b):
| $50\,000r^{n-1} > 200\,000$ | M1 | Allow equals sign or wrong inequality sign; condone slips |
| $r^{n-1} > 4 \Rightarrow (n-1)\log r > \log 4$ | M1 | Introducing logs and dealing correctly with the power |
| $n > \frac{\log 4}{\log r} + 1$ | A1cso | |
### Part (c):
| $r = 1.09$: $n > \frac{\log 4}{\log 1.09} + 1$ or $n - 1 > \frac{\log 4}{\log 1.09}$ ($n > 17.086...$) | M1 | Allow equality |
| Year 18 or 2023 | A1 | If one of these is correct, ignore the other |
### Part (d):
| $S_n = \frac{a(1-r^n)}{1-r} = \frac{50000(1-1.09^{10})}{1-1.09}$ | M1 A1 | |
| £760,000 | A1 | Must be this answer, nearest £10000 |
---8. A trading company made a profit of $\pounds 50000$ in 2006 (Year 1).
A model for future trading predicts that profits will increase year by year in a geometric sequence with common ratio $r , r > 1$.
The model therefore predicts that in 2007 (Year 2) a profit of $\pounds 50000 r$ will be made.
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the predicted profit in Year $n$.
The model predicts that in Year $n$, the profit made will exceed $\pounds 200000$.
\item Show that $n > \frac { \log 4 } { \log r } + 1$.
Using the model with $r = 1.09$,
\item find the year in which the profit made will first exceed $\pounds 200000$,
\item find the total of the profits that will be made by the company over the 10 years from 2006 to 2015 inclusive, giving your answer to the nearest $\pounds 10000$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q8 [9]}}