Mitchell took out a loan for $1100 at a 19.2% APR, compounded monthly, to buy a scanner. If he will make monthly payments of $71.50 to pay off the loan, how many total payments will he have to make?

Question

PLEASE HELP ??? :))

Answer 1

The easiest way to do this is to use an iterative process.

P[n]=P[n-1](1+r)-m

where P is the current balance on the loan so that P[0]=$1100, n=month number starting at zero, m=monthly payment, r=monthly rate.

So m=$71.50, r=19.2/12%=1.6%=0.016.

Therefore P[n]=1.016P[n-1]-71.50.

The table below shows the decreasing balance:

n	Balance ($)
0	1100.00
1	1046.10
2	991.34
3	935.70
4	879.17
5	821.74
6	763.38
7	704.10
8	643.86
9	582.67
10	520.49
11	457.32
12	393.13
13	327.92
14	261.67
15	194.36
16	125.97
17	56.48

At month 17 the balance is less than the monthly payment, so there are 18 payments in all, but the last payment is reduced to $56.48.

Algebraically the balance is P[n]=P(1+r)ⁿ-m((1+r)ⁿ-1)/r where P= initial loan.

From this, n=log(m/(m-rP))/log(1+r). Plugging in the values we get n=17.8 months. The table seems to confirm this if P[n]=0. If R=1+r, PRⁿ=m(Rⁿ-1)/r; rPRⁿ/m=Rⁿ-1; Rⁿ(1-rP/m)=1; Rⁿ=m/(m-rP).