I am working on the following problem and I was wondering if someone could help me solve it.
Chuck needs to purchase an item in 10 yrs. The item costs $\$200$ today, but its price inflates $4\%$ per year. To finance the purchase, Chuck deposits $\$20$ into an account at the beginning of each year for $6$ years. He deposits an additional $X$ at the beginning of year $4$,$5$ and $6$ to meet his goal. The effective annual interest rate is $10\%$. Calculate $X$.
The following is my attempt.
A), Chuck deposits $20$ for the first $6$ years at the beginning of the year, so at the end of year $6$ the accumulated value is an annuity due, so
$$A=20\ddot s_{\overline{6} \rceil.10}$$
B), He also deposits $X$ for three years so that his last $X$ coincides with his last $20$ deposit, this accumulated value is also an annuity due, so
$$B=X\ddot s_{\overline{3} \rceil.10}$$
C), If the present value of the item is $200$ and the price inflates $4\%$ each year until year $10$, the future value is simply
$$C=200(1.04)^{10}$$
From here I solved for $X$ using the equation $$(A+B)(1.04)^3=C$$ resulting in $X\approx 25$, but the answer is supposedly $\approx 8.92$.
I appreciate your help.
I don't know what the funny symbols mean, but as you say the item is going to cost
$$\$200(1.04)^{10} = \$296.05$$
in ten years, so that's how much money he has to make. His first six deposits will be worth
$$\$20(1.10)^{10} + \$20(1.10)^9 + \cdots + \$20(1.10)^5 = \$248.52$$
when he takes them out, so that leaves
$$\$296.05 - \$248.02 = \$47.52$$
that he has to accumulate from the other three deposits (plus the interest on them). (Already it's obvious that \$25 is way too high.) So we have
$$X(1.10^7 + 1.10^6 + 1.10^5) = \$47.52$$
Solving, $$X = \$8.91.$$