Compounding interest

Let's assume that interest is compounded monthly, where the monthly rate is r. Let m=monthly payment.

At the beginning of the first month, m is deposited and by the end of month 1, this amount has grown to m(1+r). At month 2 the second deposit is made so the total amount at the beginning of month 2 is m(1+r)+m.

At the end of month 2 the amount has accumulated to (m(1+r)+m)(1+r)=m(1+r)^2+m(1+r). After n months, the accumulated amount is m[(1+r)^n+(1+r)^(n-1)+...+(1+r)^2+(1+r)]=m(1+r)[1+(1+r)+...+(1+r)^(n-1)].

The sum in square brackets is the sum of a geometric progression: m(1+r)[(1+r)^n-1]/r.

So this expression evaluates to 10000. The number of months n=6*12=72; r=4.9/1200=0.004083

m(1+4.9/1200)[(1+4.9/1200)^72-1]/(4.9/1200)=10000.

m(1.004083)(0.34098)*244.898=10000; m=119.27, monthly payment.

If r is the quarterly rate, then r=4.9/400=0.01225, n=24.

28.0446m=10000, so m=356.57. This is a quarterly payment, so the monthly payment is 118.89.

by Top Rated User (1.0m points)