Originally Posted by

**tbl9301** Zack wishes to accumulate $50,000 in a fund at the end of 20 years. If he deposits $100 + x in the fund at the end of every 3 months for the first 10 years and $100 in the fund at the end of every 3 months for the second 10 years, find x to the nearest dollar if i(4) = .04.

I think it's set up something like this: (100 + x)(a angle 40)(0.01) + (100)(a angle 40)(0.01) = 50000 , which equals (100 + x)((1-v^40)/.01) + 100((1-v^40))/.01) = 50000 with v=1/1.01

... but the answer is $520 so I might be totally off. Any help would be great thanks!