Questions: How much money will there be in an account at the end of 10 years if 5000 is deposited at 5% interest compounded semiannually? (Assume no withdrawals are made.)

Solution Steps

To solve this problem, we need to use the formula for compound interest. The formula is:

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

where:

• \( A \) is the amount of money accumulated after n years, including interest.
• \( P \) is the principal amount (the initial amount of money).
• \( r \) is the annual interest rate (decimal).
• \( n \) is the number of times that interest is compounded per year.
• \( t \) is the time the money is invested for in years.

Given:

• \( P = 5000 \)
• \( r = 0.05 \)
• \( n = 2 \) (since the interest is compounded semiannually)
• \( t = 10 \)

We will plug these values into the formula to find \( A \).

We are given the following values for the compound interest formula:

• Principal amount \( P = 5000 \)
• Annual interest rate \( r = 0.05 \)
• Number of times interest is compounded per year \( n = 2 \) (semiannually)
• Time in years \( t = 10 \)

The formula for compound interest is given by:

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

Substituting the known values into the formula:

\[ A = 5000 \left(1 + \frac{0.05}{2}\right)^{2 \times 10} \]

First, we calculate the term inside the parentheses:

\[ 1 + \frac{0.05}{2} = 1 + 0.025 = 1.025 \]

Next, we calculate the exponent:

\[ nt = 2 \times 10 = 20 \]

Now we can compute \( A \):

\[ A = 5000 \times (1.025)^{20} \]

Calculating \( (1.025)^{20} \):

\[ (1.025)^{20} \approx 1.806111234669 \]

Finally, we find \( A \):

\[ A \approx 5000 \times 1.806111234669 \approx 9030.556173345 \]

Rounding to two decimal places, we have:

\[ A \approx 9030.56 \]

Final Answer