1. **Problem statement:** Hector needs 17000 in 8 years. We want to find the quarterly deposit amount he should make at the end of each quarter to reach this amount. 2. **Formula used:** For an ordinary annuity (deposits at the end of each period), the future value $FV$ is given by: $$FV = P \times \frac{(1 + r)^n - 1}{r}$$ where $P$ is the periodic deposit, $r$ is the interest rate per period, and $n$ is the total number of periods. 3. **Important rules:** - The interest rate per period $r$ is the annual nominal rate divided by the number of compounding periods per year. - The total number of periods $n$ is the number of years multiplied by the number of compounding periods per year. 4. **Given data:** - $FV = 17000$ - Number of years = 8 - Compounding quarterly means 4 periods per year ### Part (a): 6% compounded quarterly 5. Calculate $r$ and $n$: $$r = \frac{6\%}{4} = 0.06 / 4 = 0.015$$ $$n = 8 \times 4 = 32$$ 6. Substitute into the formula and solve for $P$: $$17000 = P \times \frac{(1 + 0.015)^{32} - 1}{0.015}$$ 7. Calculate $(1 + 0.015)^{32}$: $$ (1.015)^{32} \approx 1.601032 $$ 8. Calculate numerator: $$1.601032 - 1 = 0.601032$$ 9. Substitute back: $$17000 = P \times \frac{0.601032}{0.015} = P \times 40.0688$$ 10. Solve for $P$: $$P = \frac{17000}{40.0688}$$ $$P = \frac{\cancel{17000}}{\cancel{40.0688}}$$ $$P \approx 424.32$$ ### Part (b): 4% compounded quarterly 11. Calculate $r$ and $n$: $$r = \frac{4\%}{4} = 0.04 / 4 = 0.01$$ $$n = 8 \times 4 = 32$$ 12. Substitute into the formula and solve for $P$: $$17000 = P \times \frac{(1 + 0.01)^{32} - 1}{0.01}$$ 13. Calculate $(1 + 0.01)^{32}$: $$ (1.01)^{32} \approx 1.374 \text{ (rounded)} $$ 14. Calculate numerator: $$1.374 - 1 = 0.374$$ 15. Substitute back: $$17000 = P \times \frac{0.374}{0.01} = P \times 37.4$$ 16. Solve for $P$: $$P = \frac{17000}{37.4}$$ $$P = \frac{\cancel{17000}}{\cancel{37.4}}$$ $$P \approx 454.01$$ **Final answers:** - (a) Quarterly deposit at 6%: approximately 424.32 - (b) Quarterly deposit at 4%: approximately 454.01