1. We solve by the formula of geometric progression:
"Sn=\\frac{b1(q^n-1)}{q-1}"
Sn=$318,921.46
b1=200
q=1,05
"318921.46=\\frac{200(1.05^n-1)}{1.05-1}"
"318921.46=4000(1.05^n-1)"
"1.05^n-1=79.73"
"1.05^n=80.73"
n=log1.0580.73
n=89,99999
90 payments are necessary to pay off the loan amount assuming no deposit was made
2.Find the last member of the progression:
"bn=b1\\times1.05^{n-1}"
"bn=200\\times1.05^{90-1}"
"bn=200\\times1.05^{89}"
bn=15 377.21=15378
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