Questions: Complete the table and find the balance A if 3900 is invested at an annual percentage rate of 3% for 10 years and compounded n times a year. n 1 2 4 12 365 A Complete the table. n 1 2 4 12 365 A (Do not round until the final answer. Then round to the nearest cent as needed.)

Solution Steps

Let \( P = 3900 \) (the principal amount), \( r = 0.03 \) (the annual interest rate), and \( t = 10 \) (the time in years). We will calculate the balance \( A \) for different compounding frequencies \( n \).

The balance \( A \) after \( t \) years when compounded \( n \) times a year is given by the formula:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

We will compute \( A \) for the following values of \( n \):

• \( n = 1 \)
• \( n = 2 \)
• \( n = 4 \)
• \( n = 12 \)
• \( n = 365 \)

For each \( n \), substitute the values into the formula:

1. For \( n = 1 \): \[ A = 3900 \left(1 + \frac{0.03}{1}\right)^{1 \cdot 10} = 3900 \left(1.03\right)^{10} \]

2. For \( n = 2 \): \[ A = 3900 \left(1 + \frac{0.03}{2}\right)^{2 \cdot 10} = 3900 \left(1.015\right)^{20} \]

3. For \( n = 4 \): \[ A = 3900 \left(1 + \frac{0.03}{4}\right)^{4 \cdot 10} = 3900 \left(1.0075\right)^{40} \]

4. For \( n = 12 \): \[ A = 3900 \left(1 + \frac{0.03}{12}\right)^{12 \cdot 10} = 3900 \left(1.0025\right)^{120} \]

5. For \( n = 365 \): \[ A = 3900 \left(1 + \frac{0.03}{365}\right)^{365 \cdot 10} = 3900 \left(1.00008219178\right)^{3650} \]

After calculating \( A \) for each \( n \), round the results to the nearest cent. The final balances are:

• For \( n = 1 \): \( A \approx 5241.27 \)
• For \( n = 2 \): \( A \approx 5252.73 \)
• For \( n = 4 \): \( A \approx 5258.56 \)
• For \( n = 12 \): \( A \approx 5262.48 \)
• For \( n = 365 \): \( A \approx 5264.38 \)

Final Answer