The loan amount is 109000-21800=$87,200. This is the amount Angela must finance.

Let P=$87,200 for the sake of illustrating the formulas. We’ll call the monthly payment m and the monthly rate of interest r=0.037/12=0.003083 approx. Let n be the repayment period so n=25×12=300 months. Now we can create an algebraic expression for what happens in the first month. After one month the loan amount gains interest=rP ($268.87 in figures). So the loan with one month’s interest is P+rP=P(1+r). Then the monthly payment (which is yet to be found) is deducted: P(1+r)-m, which becomes the amount left to pay at the beginning of the second month. This gains interest as before so if P₁=P₁(1+r), then P₁=(P(1+r)-m)(1+r).

Now we deduct the second payment: P(1+r)-m)(1+r)-m. This is the amount to pay at the beginning of the third month. After another month’s payment the formula becomes (P(1+r)-m)(1+r)-m)(1+r)-m.

We end up with a series: P(1+r)³-m(1+r)²-m(1+r)-m for the amount left to pay after 3 months.

If we extend the period to n months, we get P(1+r)ⁿ-m(1+(1+r)+(1+r)²+...+(1+r)ⁿ⁻¹).

After n months there is no more to pay so this expression equals zero, and:

m(1+(1+r)+(1+r)²+...+(1+r)ⁿ⁻¹)=P(1+r)ⁿ.

The left hand side reduces to a formula for the sum of the terms of a geometric progression:

m((1+r)ⁿ-1)/r=P(1+r)ⁿ and from this m can be calculated:

Let X=(1+0.037/12)³⁰⁰=2.151828 approx, the growth factor.

m=87200X÷((X-1)/(0.037/12))=219594.18/492.42=$445.95.

P(1+r)ⁿ is the future value of the loan=87200X=$219,594.18 approx.

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