| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| 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 with clear context and standard formulas. Part (a) requires simple calculation with r=1.08, part (b) involves solving a basic geometric equation using logarithms, and part (c) applies the sum formula directly. All techniques are routine for A-level students who have learned geometric sequences, requiring no problem-solving insight beyond recognizing the GP structure. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_3 = £20000 \times 1.08^2 = (£)23328\) | B1* | Show that question - method must be seen. Condone missing units. E.g. \(£20000 \times 1.08^2\) or \(£20000 \times 108\% \times 108\%\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20000 \times 1.08^{n-1} > 65000\) | M1 | Sets up inequality/equation to allow problem to be solved. Condone slips on 20000 and 65000 but 1.08 must be correct |
| \(1.08^{n-1} > \frac{13}{4} \Rightarrow n-1 > \frac{\ln(3.25)}{\ln(1.08)}\) or \(n-1 > \log_{1.08}\left(\frac{13}{4}\right)\) | M1 | Correct strategy using logs. Condone slips on 20000, 65000, and error on 1.08 |
| Year 17 | A1 | Interprets decimal value and gives correct year number |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{20} = \frac{20000(1-1.08^{20})}{1-1.08}\) | M1 | Attempts correct sum formula for 20 years. Condone slip on 20000 or 1.08. Do not condone 20 appearing as 19 |
| Awrt \((£)\ 915000\) | A1 | Condone missing unit. Answer without working scores M0 A0 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_3 = £20000 \times 1.08^2 = (£)23328$ | B1* | Show that question - method must be seen. Condone missing units. E.g. $£20000 \times 1.08^2$ or $£20000 \times 108\% \times 108\%$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20000 \times 1.08^{n-1} > 65000$ | M1 | Sets up inequality/equation to allow problem to be solved. Condone slips on 20000 and 65000 but 1.08 must be correct |
| $1.08^{n-1} > \frac{13}{4} \Rightarrow n-1 > \frac{\ln(3.25)}{\ln(1.08)}$ or $n-1 > \log_{1.08}\left(\frac{13}{4}\right)$ | M1 | Correct strategy using logs. Condone slips on 20000, 65000, and error on 1.08 |
| Year 17 | A1 | Interprets decimal value and gives correct year number |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{20} = \frac{20000(1-1.08^{20})}{1-1.08}$ | M1 | Attempts correct sum formula for 20 years. Condone slip on 20000 or 1.08. Do not condone 20 appearing as 19 |
| Awrt $(£)\ 915000$ | A1 | Condone missing unit. Answer without working scores M0 A0 |
---\begin{enumerate}
\item In this question you should show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
A company made a profit of $\pounds 20000$ in its first year of trading, Year 1\\
A model for future trading predicts that the yearly profit will increase by $8 \%$ each year, so that the yearly profits will form a geometric sequence.
According to the model,\\
(a) show that the profit for Year 3 will be $\pounds 23328$\\
(b) find the first year when the yearly profit will exceed £65000\\
(c) find the total profit for the first 20 years of trading, giving your answer to the nearest £1000
\hfill \mbox{\textit{Edexcel Paper 1 2021 Q5 [6]}}