Let’s use algebra to find a general solution to the problem.
Let P=initial amount borrowed, m=monthly payment, T=number of years, r=monthly rate.
Later we can plug in actual values.
Now let’s start with the amount P. After a month this has grown to P(1+r). Then the first monthly payment is made, reducing the loan to P(1+r)-m.
One month later this reduced amount grows to (P(1+r)-m)(1+r) and is reduced by the monthly payment to (P(1+r)-m)(1+r)-m. A month later the reduced amount becomes ((P(1+r)-m)(1+r)-m)(1+r)-m.
Let’s expand this: P(1+r)³-m(1+(1+r)+(1+r)²).
We want the loan to reduce to zero after n years =12n months.
If we carry on applying the growth and the monthly payments we end up after 12n months with:
P(1+r)¹²ⁿ-m(1+(1+r)+(1+r)²+...+(1+r)¹²ⁿ⁻¹)=0.
It seems we have a geometric series for m, the sum of which is m((1+r)¹²ⁿ-1)/(1+r-1).
Therefore P(1+r)¹²ⁿ=m((1+r)¹²ⁿ-1)/r.
We have a general formula. Let’s plug in some values.
What is P? $35000 has been borrowed but payments don’t start for 21 months. The annual rate is 4.8% so the monthly rate r=0.4% which is 0.004. P=35000(1.004)²¹=$38060.63 approx.
We also need 12n=12×20=240 months. And we need 1.004²⁴⁰=2.6067 approx.
m(2.6067-1)/0.004=38060.63×2.6067.
1.6067m=0.004×38060.63×2.6067=39.685 and m=39.685/1.6067=$24.70.
The monthly payment is therefore $24.70.