A student takes out a loan for 30,000. The annual nominal interest rate is 9%, convertible semiannually. The student pays off the loan in five years with monthly payments beginning one month from today. The first payment is 500, and each subsequent payment is X more than the previous payment.

Determine which of the following is an equation of value that can be used to solve for X.

• [[math]]\quad 30,000=\sum_{n=0}^{60} \frac{500+n X}{(1.015)^{\frac{n}{2}}}[[/math]]
• [[math]]30,000=\sum_{n=0}^{60} \frac{500+n X}{(1.045)^{\frac{n}{6}}}[[/math]]
• [[math]]30,000=\sum_{n=1}^{60} \frac{500+n X}{(1.045)^{\frac{n}{6}}}[[/math]]
• [[math]]30,000=\sum_{n=1}^{60} \frac{500+(n-1) X}{(1.015)^{\frac{n}{2}}}[[/math]]
• [[math]]\quad 30,000=\sum_{n=1}^{60} \frac{500+(n-1) X}{(1.045)^{\frac{n}{6}}}[[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

Let n index the payment times in months. Then for a payment at time n, the discount factor is

[[math]]\frac{1}{(1.045)^{n / 6}}.[[/math]]

The end-of-month payment at month [math]n[/math] is be [math]500+(n-1) X[/math]. Therefore, we have

[[math]]30,000=\sum_{n=1}^{60} \frac{500+(n-1) X}{(1.045)^{n / 6}}.[[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.