Question

Allysha just borrowed 35,800 dollars. She plans to repay this loan by making a special payment of 6,400 dollars in 5 years and by making regular annual payments of 6,300 dollars per year until the loan is paid off. If the interest rate on the loan is 8.2 percent per year and she makes her first regular annual payment of 6,300 dollars in one year, then how many regular annual payments of 6,300 dollars must Allysha make? Round your answer to 2 decimal places (for example, 2.89, 14.70, or 6.00).

Answer 1

35800=6400/1.082^5+6300/1.082+6300/1.082^2+.........+6300/1.082^n[where n=time period]

35800=(6400*0.674316359)+6300[1/1.082+1/1.082^2+..............+1/1.082^n]

35800=4315.624698+6300[1/1.082+1/1.082^2+..............+1/1.082^n]

(35800-4315.624698)/6300=[1/1.082+1/1.082^2+..............+1/1.082^n]

[1/1.082+1/1.082^2+..............+1/1.082^n]=4.997519889

Hence a=first term=1/1.082

r=common ratio=2nd term/1st term

=(1/1.082^2)/1/1.082

=1/1.082

Sum of geomteric progression=a(1-r^n)/(1-r)

=[(1/1.082)(1-(1/1.082)^n)]/[1-(1/1.082)]

=[0.924214417(1-0.924214417^n)]/[1-0.924214417]

=[0.924214417(1-0.924214417)]/0.075785583

=12.19512182(1-0.924214417^n)

Hence

4.997519889=12.19512182(1-0.924214417^n)

(4.997519889/12.19512182)=(1-0.924214417^n)

(1-0.924214417^n)=0.409796635

1-0.409796635=0.924214417^n

0.924214417^n=0.590203365

Taking log on both sides;

n*log0.924214417= log 0.590203365

n= log 0.590203365/log0.924214417

=6.69 years(Approx).