# Annuities question - Equations of value

Chuck needs to purchase an item in 10 years. The item costs 200 today, but its price inflates at 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 years 4, 5, and 6 to meet his goal. The annual effective interest rate is 10%. Calculate X.

This is how i interpret the problem: You have 5 cash flows starting from 0 to 5 of $20. You also have 3 cash flows at t=4,5,6. I used annuity due formula to shift former cash flow to year 6, and then accumulate it to year 10 by the 4 remaining years. I used the same approach for the latter: But this does not give me the right answer. Can someone please tell me what I'm doing wrong? Thanks in advance. ## 2 Answers These are the steps I'd follow:$200 today times (1.04)^10 = Cost in year 10.

The 6 deposits of \$20 will be one time value calculation with a resulting year 7 final value. You then must apply 10% for 3 years (1.1)^3 to get the 10th year result.

You now have the shortfall. Divide that by the same (1.1)^3 to shift the present value to start of year 7. (this step might confuse you?)

You are left with a problem needing 3 same deposits, a known rate, and desired FV. Solve from there.

(Also, welcome from quant.SE. This site doesn't support LATEX, so I edited the image above.)

The solution is x = 8.92.

This assumes that Chuck's six years of deposits start from today, so that the first deposit accumulates 10 years of gain, i.e. 20*(1 + 0.1)^10. The second deposit gains nine years' interest: 20*(1 + 0.1)^9 and so on ... If you want to do this calculation using the formula for an annuity due, i.e.

http://www.financeformulas.net/Future-Value-of-Annuity-Due.html where you have to bear in mind this is for the whole time span (k = 1 to n), so for just the first six years you need to calculate for all ten years then subtract another annuity calculation for the last four years. So the full calculation is: As you can see it's not very neat, because the standard formula is for a whole time span. You could make it a little tidier by using a formula for k = m to n instead, i.e. So the calculation becomes which can be done with simple arithmetic (and doesn't actually need a solver).