Use of annuities
Example 7
Jessica takes out a car loan for $\$ 24000$ with an interest rate of $10 \%$ p.a. compound interest. She agrees to a regular yearly payment of $\$ 5500$ over six years. A table for this annuity is shown below.
Determine, to the nearest cent:
a the interest paid when payment 1 is received 2400.00
b the principal reduction when payment 3 is received 3751.00
c the balance of the annuity after payment 4 has been received 9612.90
d the value of the last payment if the balance of the annuity is to be zero after the 6th payment is received
e the total amount of interest paid.

Step 1 ：\(A = P(1 + \frac{r}{n})^{nt}\)

Step 2 ：\(A_1 = 24000(1 + \frac{0.1}{1})^{1} \approx 26400\)

Step 3 ：\(I_1 = A_1 - 24000 = 2400\)

Step 4 ：\(\boxed{2400.00}\)

Step 5 ：\(A_3 = 24000(1 + \frac{0.1}{1})^{3} - 5500(1 + \frac{0.1}{1})^{2} - 5500(1 + \frac{0.1}{1})^{1} \approx 13749\)

Step 6 ：\(P_3 = 5500 - 3751 = 1749\)

Step 7 ：\(\boxed{3751.00}\)

Step 8 ：\(A_4 = 24000(1 + \frac{0.1}{1})^{4} - 5500(1 + \frac{0.1}{1})^{3} - 5500(1 + \frac{0.1}{1})^{2} - 5500(1 + \frac{0.1}{1})^{1} \approx 9612.90\)

Step 9 ：\(\boxed{9612.90}\)

Step 10 ：\(A_6 = 24000(1 + \frac{0.1}{1})^{6} - 5500(1 + \frac{0.1}{1})^{5} - 5500(1 + \frac{0.1}{1})^{4} - 5500(1 + \frac{0.1}{1})^{3} - 5500(1 + \frac{0.1}{1})^{2} - 5500(1 + \frac{0.1}{1})^{1} = 0\)

Step 11 ：\(P_6 = 5500 - 0 = 5500\)

Step 12 ：\(\boxed{5500}\)

Step 13 ：\(I_{total} = 6 \cdot 5500 - 24000 = 33000 - 24000 = 9000\)

Step 14 ：\(\boxed{9000}\)