| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric sequences requiring only direct formula substitution. Part (a) is a 'show that' verification, part (b) uses logarithms in a standard depreciation context, and parts (c)-(d) apply the GP sum formula to a routine compound interest problem. All techniques are standard C2 material with no problem-solving insight required. |
| 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 |
|---|---|---|
| \(18000 \times (0.8)^3 = £9216\) | B1cso (1) | Answer printed so need working; may see \(\frac{4}{5}\) or 80%; may show 14400, 11520, 9216 |
| Answer | Marks | Guidance |
|---|---|---|
| \(18000 \times (0.8)^n < 1000\) | M1 | 1st M1 for attempt to use \(n\)th term and 1000; allow \(n\) or \(n-1\), allow \(>\) or \(=\) |
| \(n\log(0.8) < \log\left(\frac{1}{18}\right)\) | M1 | 2nd M1 for use of logs to find \(n\); allow \(>\) or \(=\) |
| \(n > \frac{\log\left(\frac{1}{18}\right)}{\log(0.8)} = 12.952\ldots\), so \(n = 13\) | A1cso (3) | A1 needs \(n=13\); must follow both M marks; condone slips in inequality signs |
| Answer | Marks | Guidance |
|---|---|---|
| \(u_5 = 200 \times (1.12)^4 = £314.70\) or \(£314.71\) | M1, A1 (2) | M1 for use of their \(a\) and \(r\) in formula for 5th term; NB \(314.7 =\) A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{15} = \frac{200(1.12^{15}-1)}{1.12-1}\) or \(\frac{200(1-1.12^{15})}{1-1.12} = 7455.94\ldots\) awrt £7460 | M1A1, A1 (3) | M1 for use of sum formula with their \(a\) and \(r\), using \(n\) not \(n-1\); 1st A1 for fully correct expression; adds 15 terms \(200+224+250.88+\ldots+(977.42)\) is M1; seeing 977… is A1 |
## Question 6:
### Part (a):
| $18000 \times (0.8)^3 = £9216$ | B1cso (1) | Answer printed so need working; may see $\frac{4}{5}$ or 80%; may show 14400, 11520, 9216 |
### Part (b):
| $18000 \times (0.8)^n < 1000$ | M1 | 1st M1 for attempt to use $n$th term and 1000; allow $n$ or $n-1$, allow $>$ or $=$ |
| $n\log(0.8) < \log\left(\frac{1}{18}\right)$ | M1 | 2nd M1 for use of logs to find $n$; allow $>$ or $=$ |
| $n > \frac{\log\left(\frac{1}{18}\right)}{\log(0.8)} = 12.952\ldots$, so $n = 13$ | A1cso (3) | A1 needs $n=13$; must follow both M marks; condone slips in inequality signs |
### Part (c):
| $u_5 = 200 \times (1.12)^4 = £314.70$ or $£314.71$ | M1, A1 (2) | M1 for use of their $a$ and $r$ in formula for 5th term; NB $314.7 =$ A0 |
### Part (d):
| $S_{15} = \frac{200(1.12^{15}-1)}{1.12-1}$ or $\frac{200(1-1.12^{15})}{1-1.12} = 7455.94\ldots$ awrt £7460 | M1A1, A1 (3) | M1 for use of sum formula with their $a$ and $r$, using $n$ not $n-1$; 1st A1 for fully correct expression; adds 15 terms $200+224+250.88+\ldots+(977.42)$ is M1; seeing 977… is A1 |
---6. A car was purchased for $\pounds 18000$ on 1 st January.
On 1st January each following year, the value of the car is $80 \%$ of its value on 1st January in the previous year.
\begin{enumerate}[label=(\alph*)]
\item Show that the value of the car exactly 3 years after it was purchased is $\pounds 9216$.
The value of the car falls below $\pounds 1000$ for the first time $n$ years after it was purchased.
\item Find the value of $n$.
An insurance company has a scheme to cover the maintenance of the car. The cost is $\pounds 200$ for the first year, and for every following year the cost increases by $12 \%$ so that for the 3rd year the cost of the scheme is $\pounds 250.88$
\item Find the cost of the scheme for the 5th year, giving your answer to the nearest penny.
\item Find the total cost of the insurance scheme for the first 15 years.\\
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q6 [9]}}