To find the average amount earned per year between the 5th year and the 10th year for an investment of $5000 compounded continuously at 4%, we will use the formula for continuous compounding:
\[ A(t) = P e^{rt} \]
where:
- \( A(t) \) is the amount of money accumulated after \( t \) years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( e \) is Euler's number (approximately 2.71828).
- \( t \) is the time in years.
For this problem:
- \( P = 5000 \)
- \( r = 0.04 \)
The formula becomes:
\[ A(t) = 5000 e^{0.04t} \]
Now, we will calculate the amount at \( t = 5 \) and \( t = 10 \):
-
Calculate \( A(5) \): \[ A(5) = 5000 e^{0.04 \times 5} = 5000 e^{0.2} \] Using \( e^{0.2} \approx 1.2214 \): \[ A(5) \approx 5000 \times 1.2214 \approx 6107.00 \]
-
Calculate \( A(10) \): \[ A(10) = 5000 e^{0.04 \times 10} = 5000 e^{0.4} \] Using \( e^{0.4} \approx 1.4918 \): \[ A(10) \approx 5000 \times 1.4918 \approx 7459.00 \]
-
Find the total amount earned between years 5 and 10: \[ \text{Total amount earned from year 5 to year 10} = A(10) - A(5) \] \[ \text{Total amount earned} = 7459.00 - 6107.00 = 1352.00 \]
-
Calculate the average amount earned per year between the 5th year and the 10th year: The duration from year 5 to year 10 is 5 years. Thus, \[ \text{Average amount earned per year} = \frac{\text{Total amount earned}}{\text{Number of years}} = \frac{1352.00}{5} = 270.40 \]
Thus, the average amount earned per year between the 5th year and the 10th year is $270.40.
1 answer