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).