Money is invested in a savings account with a nominal interest rate of 6% convertible...?
The first thing that you need to do is convert the nominal interest rate into an annual effective rate, so that the periods for interest compounding and inflation are the same: (1+.06/12)^12=1+i 1.005^12=1+i i=.0617=6.17% effective annual. Then there are two ways to proceed. I will show both in case one of them makes more sense to you: 1. Recall the formula i(real)=(i-r)/(1+r), where i is the "nominal" (not real) periodic interest rate and r is the periodic inflation rate. From this formula, we get 1+i(real)=(1+i)/(1+r). In this problem, i(real) is the real effective rate of interest per year, so we get [1+i(real)]^3= (1.0617/1.034) * (1.0617/1.062) * (1.0617/1.062). Then i(real)=.0087 = .87% effective annual. 2. You can also compute the real annual effective interest rate each year, then compound over the 3-year period. Recall that i(real)=(i-r)/(1+r). So i(real 1) = (.0617-.034)/1.034 = .0268 = 2.68% i(real 2) = i(real 3) = (.0617-.062)/1.062 = -.0003 = -.03% So [1+i(real)]^3 = [1+i(real 1)]*[1+i(real 2)]*[1+i(real 2)] [1+i(real)]^3 = (1.0268)(1 - .0003)(1 - .0003) [1+i(real)]^3 = 1.0268*.9997^2 i (real) = .0087 = .87% effective annual.