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M
monthly repayments
in your eqn above there is a G.S. (a=1, r=1.006, n=n); and you solve for An = 0 Sn = [{(1.006^n)-1}/0.006] .: 2800[(1.006^n)-1] = 1200(1.006^n) ==> 2800(1.006^n) - 2800 = 1200(1.006^n) ==> 1600(1.006^n) = 2800 ==> 1.006^n = 1.75 ==> ln(1.006^n) = ln 1.75 ==> n ln (1.006) = ln 1.75...- mbenson
- Post #22
- Forum: Mathematics
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