Trish had a loan with a balance of 4000 at the beginning of month 1. Starting with month 1, and every month thereafter, she made a payment of X in the middle of the month. At the beginning of month 4, and every 6 months thereafter, she borrowed an additional 800. Trish’s loan balance became 4000 again at the end of month 36. The annual nominal interest rate for the loan is 26.4%, convertible quarterly.

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

• [[math]]\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{2n-1}}}=\sum_{n=1}^{36}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{n-0.5}{3}}}} [[/math]]
• [[math]]\sum_{n=1}^{6}\frac{800}{\left(1+\frac{0.2640}{12}\right)^{6n-3}}=\sum_{n=1}^{36}\frac{X}{\left(1+\frac{0.2640}{12}\right)^{n-0.5}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{6n-2}{3}}}}+\sum_{n=1}^{\frac{\displaystyle4000}{4}}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{\frac{n}{3}}}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{12}}\right)^{6n-3}}}={\frac{4000}{\left(1+{\frac{0.2640}{12}}\right)^{36}}}+\sum_{n=1}^{36}{\frac{X}{\left(1+{\frac{0.2640}{12}}\right)^{n-0.5}}} [[/math]]
• [[math]]4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.2640}{4}}\right)^{2n-1}}}={\frac{4000}{\left(1+{\frac{0.2640}{4}}\right)^{12}}}+\sum_{n=1}^{\infty}{\frac{X}{\left(1+{\frac{0.2640}{4}}\right)^{n-0.5}}} [[/math]]

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

Solution: E

The present value of the payments (4000 at month 36 plus the payments of X) must match the present value of the present value of the amounts borrowed (4000 at month 0 plus the payments of 800).

The quarterly interest rate is 0.264/4 and all payment times should be in quarters of a year. On that time scale, the 4000 at month 36 is at time 12. The payments of 4000 are at times 1/6, 3/6, 5/6, ..., 71/6 and there are 36 such payments. One way to write the present value of these payments is

[[math]] \frac{4000}{\left(1\!\!+\!\frac{0.264}{4}\right)^{12}} + \sum_{n=1}^{36}\frac{X}{\left(1\!+\!\frac{0.264}{4}\right)^{\frac{n-0.5}{3}}}. [[/math]]

The payments of 800 are at times 1, 3, 5, 7, 9, and 11, in quarters. One way to write the present value of these payments plus the initial debt of 4000 is

[[math]] 4000+\sum_{n=1}^{6}{\frac{800}{\left(1+{\frac{0.264}{12}}\right)^{2n-1}}}. [[/math]]

These are the two sides of equation in answer choice E.

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