| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | January |
| Marks | 10 |
| 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) involves calculating two terms and finding their difference, part (b) requires solving a simple inequality using logarithms, and part (c) applies the standard GP sum formula. All techniques are routine for C1/C2 level with no problem-solving insight needed. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(275000\times(1.1)^5\) or finds £442890.25 or uses \(275000\times(1.1)^4\) or finds £402627.50 | M1 | |
| Finds both and subtracts to give £40262.75 and concludes approx. £40300 | M1 A1* | |
| Or: \(275000\times(1.1)^5 - 275000\times(1.1)^4 = \text{awrt } 40260 = 40300\ (3\text{sf})\) | M1 M1 A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Puts \(275000\times(1.1)^{n-1}>1000000\) or \(275000\times(1.1)^{n-1}=1000000\) | M1 | |
| \((1.1)^{n-1}>\frac{1000000}{275000}\) (or \(\frac{40}{11}\) or 3.63 or 3.64) | M1 | |
| \(n-1>\frac{\log\left(\frac{40}{11}\right)}{\log 1.1}\) or \(n-1=\frac{\log\left(\frac{40}{11}\right)}{\log 1.1}\) | M1 | |
| \((n>14.5\) or \(n>14.6\) or \(n=15)\) so the year is 2030 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(S=\frac{275000(1.1^n-1)}{1.1-1}\) or \(S=\frac{275000(1-1.1^n)}{1-1.1}\) | M1 | |
| Uses \(n=11\) in formula | A1 | |
| Awrt £5 096 100 | A1 |
## Question 12:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $275000\times(1.1)^5$ or finds £442890.25 or uses $275000\times(1.1)^4$ or finds £402627.50 | M1 | |
| Finds both and subtracts to give £40262.75 and concludes approx. £40300 | M1 A1* | |
| Or: $275000\times(1.1)^5 - 275000\times(1.1)^4 = \text{awrt } 40260 = 40300\ (3\text{sf})$ | M1 M1 A1* | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Puts $275000\times(1.1)^{n-1}>1000000$ or $275000\times(1.1)^{n-1}=1000000$ | M1 | |
| $(1.1)^{n-1}>\frac{1000000}{275000}$ (or $\frac{40}{11}$ or 3.63 or 3.64) | M1 | |
| $n-1>\frac{\log\left(\frac{40}{11}\right)}{\log 1.1}$ or $n-1=\frac{\log\left(\frac{40}{11}\right)}{\log 1.1}$ | M1 | |
| $(n>14.5$ or $n>14.6$ or $n=15)$ so the year is 2030 | A1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $S=\frac{275000(1.1^n-1)}{1.1-1}$ or $S=\frac{275000(1-1.1^n)}{1-1.1}$ | M1 | |
| Uses $n=11$ in formula | A1 | |
| Awrt £5 096 100 | A1 | |12. A business is expected to have a yearly profit of $\pounds 275000$ for the year 2016. The profit is expected to increase by $10 \%$ per year, so that the expected yearly profits form a geometric sequence with common ratio 1.1
\begin{enumerate}[label=(\alph*)]
\item Show that the difference between the expected profit for the year 2020 and the expected profit for the year 2021 is $\pounds 40300$ to the nearest hundred pounds.
\item Find the first year for which the expected yearly profit is more than one million pounds.
\item Find the total expected profits for the years 2016 to 2026 inclusive, giving your answer to the nearest hundred pounds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2015 Q12 [10]}}