1. **State the problem:** Find the future value of a retirement savings account with an APR of 6.3% after 45 years, with annual contributions of 1400 made at the end of each year. 2. **Formula used:** The future value of an ordinary annuity (contributions at the end of each period) is given by: $$FV = P \times \frac{(1 + r)^n - 1}{r}$$ where: - $P$ = annual contribution - $r$ = annual interest rate (as a decimal) - $n$ = number of years 3. **Substitute values:** $$P = 1400, \quad r = 0.063, \quad n = 45$$ 4. **Calculate $(1 + r)^n$:** $$1 + r = 1 + 0.063 = 1.063$$ $$1.063^{45} \approx 13.267$$ 5. **Calculate numerator:** $$13.267 - 1 = 12.267$$ 6. **Calculate fraction:** $$\frac{12.267}{0.063} \approx 194.667$$ 7. **Calculate future value:** $$FV = 1400 \times 194.667 = 272533.8$$ 8. **Final answer:** The value of the retirement savings account after 45 years is approximately **272533.80**.