A credit card company charges an annual effective interest rate of 16.8%. Interest accumulates from the date of purchase. A borrower's credit card balance was X at the beginning of month 1. Starting with month 1, the borrower purchased satellite internet service, resulting in a 79.99 charge on the credit card at the middle of each month. The borrower paid 250 at the end of each month. Immediately after the payment in month 15, the balance was 3000.

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

• [[math]]\quad X+\sum_{n=1}^{15} \frac{79.99}{(1.014)^{n-0.5}}=\frac{3000}{(1.014)^{15}}+\sum_{n=1}^{15} \frac{250}{(1.014)^n}[[/math]]
• [[math]]\quad X+\sum_{n=1}^{16} \frac{79.99}{(1.168)^{\frac{n+0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{16} \frac{250}{(1.168)^{\frac{n}{12}}}[[/math]]
• [[math]]\quad X+\sum_{n=1}^{16} \frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}=\frac{3000}{(1.168)^{\frac{4}{3}}}+\sum_{n=1}^{16} \frac{250}{(1.168)^{\frac{n}{12}}}[[/math]]
• [[math]]\quad X+\sum_{n=1}^{15} \frac{79.99}{(1.168)^{\frac{n+0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{15} \frac{250}{(1.168)^{\frac{n}{12}}}[[/math]]
• [[math]]\quad X+\sum_{n=1}^{15} \frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{15} \frac{250}{(1.168)^{\frac{n}{12}}}[[/math]]

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

Solution: E

Note that had the borrower 1) charged X at the end of month 0, and 2) paid off the remaining 3000 at the beginning of month 16, then the initial and final balances would have become 0. In that situation, the present value of the amounts charged to the credit card, minus the present values of the payments, would have been 0 in the 15-month period.

The amounts charged to the card were 79.99 at each of times [math]t=\frac{n-0.5}{12}[/math], for each whole number of [math]n[/math] from 1 to 15 .

The monthly payments were 250 at each of times [math]t=\frac{n}{12}[/math], for each whole number [math]n[/math] from 1 to 15 inclusive.

Then to make the final balance 0 , the final (additional) payment would have been 3000 at time [math]t=\frac{15}{12}=\frac{5}{4}[/math].

Therefore, we have

[[math]] \begin{aligned} & X+\sum_{n=1}^{15} \frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}-\sum_{n=1}^{15} \frac{250}{(1.168)^{\frac{n}{12}}}-\frac{3000}{(1.168)^{\frac{5}{4}}}=0 \\ & X+\sum_{n=1}^{15} \frac{79.99}{(1.168)^{\frac{n-0.5}{12}}}=\frac{3000}{(1.168)^{\frac{5}{4}}}+\sum_{n=1}^{15} \frac{250}{(1.168)^{\frac{n}{12}}} \end{aligned} [[/math]]

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