in Other Math Topics by Level 1 User (140 points)

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

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


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).

by Top Rated User (1.0m points)

Related questions

1 answer
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,021 questions
96,296 answers
24,339 users