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


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, $$X = \$8.91.$$

  • $\begingroup$ That actually answers my question, thank you. The notation that I am using is called actuarial notation and $\ddot s_{\overline{n}\rceil i}$ is the "future value factor of annuity-due with n payments with interest i", which is equal to $(1+i)\frac{(1+i)^n-1}{i}$ $\endgroup$
    – hyg17
    Oct 18 '14 at 23:44

Since the amount $(A+B)$ grows at the rate of $10\%$ in Years 7 through 10, your final equation should be:

$(A+B)(1.10)^4 = C$

This will give you $X = 8.914$

  • $\begingroup$ Thanks for the comment, but I did not get the right answer with that adjustment. Although, I usually find it very difficult to decide how many conversion periods there are without using my fingers to count like a toddler. lol $\endgroup$
    – hyg17
    Oct 18 '14 at 23:41
  • $\begingroup$ The multiplier in A is 8.487 so A is 169.74. The multiplier in B is 3.641 so B is 3.641X. C is 296.05. Solving (A+B)(1.10)^4=296.05 gives the value of X you are seeking. $\endgroup$
    – Rao A.
    Oct 19 '14 at 22:27

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