Find the balance after 3 years if \( \$ 3200 \) is invested at \( 7 \% \) p.a. compounded half-yearly. (nearest dollar)

To find the balance after 3 years with half-yearly compounding, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] 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 or borrowed for, in years. Given: - \( P = 3200 \) - \( r = 0.07 \) - \( n = 2 \) (since it's compounded half-yearly) - \( t = 3 \) Plugging in the values: \[ A = 3200 \left(1 + \frac{0.07}{2}\right)^{2 \cdot 3} \] First, calculate \( \frac{0.07}{2} = 0.035 \). Now substitute back into the formula: \[ A = 3200 \left(1 + 0.035\right)^{6} \] \[ A = 3200 \left(1.035\right)^{6} \] Now compute \( (1.035)^{6} \): \[ (1.035)^{6} \approx 1.226 \] Now calculate \( A \): \[ A \approx 3200 \times 1.226 \approx 3923.20 \] Rounding to the nearest dollar, the balance after 3 years will be approximately \( \$3923 \).