Suppose £2500 is invested at a nominal rate of interest of 5% per annum. Calculate the amount
accumulated at the end of 10 years if interest is calculated (a) annually, (b) 4 times annually,
(c) 52 times per year, (d) continuously.

To calculate the amount accumulated after 10 years for different compounding intervals, we can use the compound interest formula for scenarios (a), (b), and (c), and the formula for continuous compounding for scenario (d).

### Compound Interest Formula:
\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]
Where:
- \( A \) = the amount accumulated after time \( t \)
- \( P \) = initial principal (£2500)
- \( r \) = nominal annual interest rate (5% = 0.05)
- \( n \) = number of compounding periods per year
- \( t \) = time in years (10 years)

### Continuous Compounding Formula:
\[
A = P e^{rt}
\]
Where:
- \( e \) = Euler's number (approximately 2.718)
- \( r \), \( P \), and \( t \) have the same meaning as in the previous formula.

Now, let's calculate the accumulated amounts for each case:

#### (a) Interest compounded **annually**:
\[
n = 1 \quad \text{(compounded once per year)}
\]
\[
A = 2500 \left(1 + \frac{0.05}{1}\right)^{1 \times 10}
\]
\[
A = 2500 \times (1.05)^{10} \approx 2500 \times 1.62889 = 4072.23
\]
The amount accumulated after 10 years with annual compounding is **£4072.23**.

#### (b) Interest compounded **4 times annually**:
\[
n = 4 \quad \text{(compounded quarterly)}
\]
\[
A = 2500 \left(1 + \frac{0.05}{4}\right)^{4 \times 10}
\]
\[
A = 2500 \times \left(1 + 0.0125\right)^{40} = 2500 \times (1.0125)^{40} \approx 2500 \times 1.64362 = 4109.05
\]
The amount accumulated after 10 years with quarterly compounding is **£4109.05**.

#### (c) Interest compounded **52 times annually**:
\[
n = 52 \quad \text{(compounded weekly)}
\]
\[
A = 2500 \left(1 + \frac{0.05}{52}\right)^{52 \times 10}
\]
\[
A = 2500 \times \left(1 + 0.0009615\right)^{520} = 2500 \times (1.0009615)^{520} \approx 2500 \times 1.64701 = 4117.53
\]
The amount accumulated after 10 years with weekly compounding is **£4117.53**.

#### (d) Interest compounded **continuously**:
\[
A = 2500 \times e^{0.05 \times 10}
\]
\[
A = 2500 \times e^{0.5} \approx 2500 \times 1.64872 = 4121.79
\]
The amount accumulated after 10 years with continuous compounding is **£4121.79**.

### Summary:
- **(a) Annually**: £4072.23
- **(b) 4 times annually (Quarterly)**: £4109.05
- **(c) 52 times annually (Weekly)**: £4117.53
- **(d) Continuously**: £4121.79

Would you like further details or clarification on any part of this?

### Related Questions:
1. How does the compounding frequency affect the accumulated amount over time?
2. What would the accumulated amount be if the investment period was 20 years?
3. How does changing the interest rate affect the final amount after 10 years?
4. What if interest was compounded monthly instead of quarterly?
5. How do different nominal interest rates impact continuous compounding?

### Tip:
Increasing the frequency of compounding periods slightly increases the final amount, but continuous compounding gives the highest return at the limit.

Suppose £2500 is invested at a nominal rate of interest of 5% per annum. Calculate the amount accumulated at the end of 10 years if interest is calculated (a) annually, (b) 4 times annually, (c) 52 times per year, (d) continuously.

Discover how to calculate the accumulated amount of £2500 over 10 years at 5% interest using various compounding methods. Learn the differences between annual, quarterly, weekly, and continuous compounding with detailed solutions. Suitable for high school students studying financial mathematics.

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