Use of annuities
Example 7
Jessica takes out a car loan for $\mathrm{\$}24000$ with an interest rate of $10\mathrm{\%}$ p.a. compound interest. She agrees to a regular yearly payment of $\mathrm{\$}5500$ 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 ：$\boxed{{\displaystyle 2400.00}}$

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 ：$\boxed{{\displaystyle 3751.00}}$

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 ：$\boxed{{\displaystyle 9612.90}}$

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 ：$\boxed{{\displaystyle 5500}}$

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

Step 14 ：$\boxed{{\displaystyle 9000}}$