| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| 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 series application with clear structure. Part (a) requires simple calculation of compound interest over 2 years, part (b) is a verification using the GP sum formula, and part (c) involves comparing monthly vs annual compounding. While it requires multiple steps and careful bookkeeping, the methods are standard C2 content with no novel insight needed—slightly easier than average due to the guided structure and verification format. |
| Spec | 1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| end of 1st year: \(500 \times 1.06 = 530\) | M1 | |
| start of 2nd year: \(530 + 500 = 1030\) | ||
| interest at end of 2nd year \(= 0.06 \times 1030 = £61.80\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| end of 8th year: \(500 \times (1.06 + 1.06^2 + \ldots + 1.06^8)\) | ||
| \(= 500 \times S_8\); GP, \(a = 1.06\), \(r = 1.06\) | B1 | |
| \(= 500 \times \frac{1.06(1.06^8 - 1)}{1.06 - 1}\) | M1 A1 | |
| \(= 5245.66 \therefore £5246\) (nearest pound) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \((1.005)^{12} = 1.0617\ldots\) | M1 A1 | |
| end of 8th year: \(500 \times \frac{1.0617(1.0617)^8 - 1]}{1.0617 - 1} = 5285.71\) | M1 A1 | |
| \(\therefore\) £40 more in account (nearest pound) | A1 | (12) |
## Question 8:
**(a)**
| Answer/Working | Marks | Notes |
|---|---|---|
| end of 1st year: $500 \times 1.06 = 530$ | M1 | |
| start of 2nd year: $530 + 500 = 1030$ | | |
| interest at end of 2nd year $= 0.06 \times 1030 = £61.80$ | M1 A1 | |
**(b)**
| Answer/Working | Marks | Notes |
|---|---|---|
| end of 8th year: $500 \times (1.06 + 1.06^2 + \ldots + 1.06^8)$ | | |
| $= 500 \times S_8$; GP, $a = 1.06$, $r = 1.06$ | B1 | |
| $= 500 \times \frac{1.06(1.06^8 - 1)}{1.06 - 1}$ | M1 A1 | |
| $= 5245.66 \therefore £5246$ (nearest pound) | A1 | |
**(c)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $(1.005)^{12} = 1.0617\ldots$ | M1 A1 | |
| end of 8th year: $500 \times \frac{1.0617(1.0617)^8 - 1]}{1.0617 - 1} = 5285.71$ | M1 A1 | |
| $\therefore$ £40 more in account (nearest pound) | A1 | **(12)** |
---8. Amy plans to join a savings scheme in which she will pay in $\pounds 500$ at the start of each year.
One scheme that she is considering pays 6\% interest on the amount in the account at the end of each year.
For this scheme,
\begin{enumerate}[label=(\alph*)]
\item find the amount of interest paid into the account at the end of the second year,
\item show that after interest is paid at the end of the eighth year, the amount in the account will be $\pounds 5246$ to the nearest pound.
Another scheme that she is considering pays $0.5 \%$ interest on the amount in the account at the end of each month.
\item Find, to the nearest pound, how much more or less will be in the account at the end of the eighth year under this scheme.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q8 [12]}}