Independent solution

How to solve this Geometrically Increasing Loan Payments question

Setup

Setup

Let X be the first year-end payment. Payment k is X times 1.06 to the k minus one.

Pk=X(1.06)k1P_k=X(1.06)^{k-1}

Model

Model

Price all ten payments at the 10% loan rate.

20,000=Xk=1101.06k11.10k20{,}000=X\sum_{k=1}^{10}\frac{1.06^{k-1}}{1.10^k}

Compute

Compute

The equation gives X = 2584.391. Value the final two payments immediately after payment 8.

B8=X(1.06)81.10+X(1.06)91.102=7,353.152B_8=\frac{X(1.06)^8}{1.10}+\frac{X(1.06)^9}{1.10^2}=7{,}353.152

Answer

Answer

The outstanding balance is approximately 7353, choice C.

B87,353(C)\boxed{B_8\approx7{,}353\quad\text{(C)}}