1. **State the problem:** Kaylynn invested an initial amount in a fund earning 2.25% annual interest compounded monthly. He starts withdrawing 5000 every six months after 4 years, and the withdrawals last for 6 years. We need to find the initial investment amount. 2. **Identify the variables:** - Annual interest rate $r = 0.0225$ - Monthly interest rate $i = \frac{0.0225}{12} = 0.001875$ - Time before withdrawals start $t_1 = 4$ years - Withdrawal period $t_2 = 6$ years - Withdrawals every 6 months (semiannual), amount $W = 5000$ - Number of withdrawals $n = \frac{6}{0.5} = 12$ 3. **Calculate the amount in the fund at the start of withdrawals:** Let initial investment be $P$. After 4 years, the amount grows to $$A = P \times (1 + i)^{12 \times 4} = P \times (1.001875)^{48}$$ 4. **Calculate the present value of withdrawals at the start of withdrawals:** Withdrawals are every 6 months, so the effective interest rate per 6 months is $$j = (1 + i)^6 - 1 = (1.001875)^6 - 1$$ Calculate $j$: $$j = 1.0113 - 1 = 0.0113$$ (approx) The present value of an annuity of $n=12$ payments of $W=5000$ at interest rate $j$ per period is $$PV = W \times \frac{1 - (1 + j)^{-n}}{j}$$ 5. **Set the amount at start of withdrawals equal to the present value of withdrawals:** $$A = PV$$ $$P \times (1.001875)^{48} = 5000 \times \frac{1 - (1.0113)^{-12}}{0.0113}$$ 6. **Calculate the right side:** Calculate $(1.0113)^{-12} = \frac{1}{(1.0113)^{12}} \approx \frac{1}{1.144} = 0.8745$ So, $$PV = 5000 \times \frac{1 - 0.8745}{0.0113} = 5000 \times \frac{0.1255}{0.0113} = 5000 \times 11.11 = 55550$$ (approx) 7. **Solve for $P$:** $$P = \frac{55550}{(1.001875)^{48}} = \frac{55550}{1.093} = 50830.47$$ (approx) **Final answer:** Kaylynn initially invested approximately **50830.47** in the fund.