Question:
30-year mortgage [n=12*30=360 months]
APR=6% compounded monthly, [i=6% APR=0.06/12=0.005 per month]
Borrowed amount=160,000, [P=160000]
Find monthly payment, A
We use the amortization formula to find the equal monthly payment
[tex]A=\frac{P(i*(1+i)^n)}{(1+i)^n-1}[/tex]
substitute values, i=0.005,n=360,P=160000
[tex]=\frac{160000(.005*(1+.005)^{360})}{(1+.005)^{360}-1}[/tex]
[tex]=\frac{160000(.005*(1.005)^{360})}{(1+.005)^{360}-1}[/tex]
[tex]=\frac{160000(.005*(1+.005)^{360})}{(1+.005)^{360}-1}[/tex]
=959.28 per month.
