Formula for Mortgage

We will let L = loan, n = number of months for repayment, starting at
end of first month, r = percentage interest rate per year, (take r/12
as monthly rate). P = amount of repayment per month (starting at end
of first month).

If we consider the loan first, this would increase by a factor
(1 + r/1200) per month, so after n months the value of the loan would
have increased to L(1 + r/1200)^n

Now consider the repayments.  These are $P per month, but the value of
the earlier repayments also increases at a compound rate (1 + r/1200). 
Thus after the second repayment, the value of the repayments is:

P + P(1 + r/1200)  and after three months it would be:
P + P(1 + r/1200) + P(1 + r/1200)^2   and after n months it would be:
P{1 + (1+r/1200) + (1+r/1200)^2 + ..... + (1+r/1200)^(n-1)}

This is a geometric series with n terms, first term = 1 and common
ratio (1+r/1200), so the sum of n terms is given by

  P{(1+r/1200)^n - 1}/{(1+r/1200)-1)}
= P{(1+r/1200)^n - 1}/(r/1200)
= 1200P{(1+r/1200)^n - 1}/r

We must now equate the total repayments to total value of the loan,
and this gives:

 1200P{(1+r/1200)^n - 1}/r  =  L(1+r/1200)^n

                    P = Lr(1+r/1200)^n/[1200{(1+r/1200)^n - 1}]

Example.  Find the monthly repayments on a loan of $20,000 over 15
years at 12 percent per year compound interest.

Here we have n = 12*15 = 180 months, r = 12, and L = 20000.
We want to find P.

1+r/1200 = 1 + 12/1200 = 1.01   and the above formula becomes


      P = {20000*12*1.01^180}/{1200*(1.01^180 - 1)}

        = {20000*12*5.99}/{1200*(5.99 - 1)}

        = 1437600/5988

        = $240.08


-Doctor Anthony,  The Math Forum
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