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).