An investor’s retirement account pays an annual nominal interest rate of 4.2%, convertible monthly. On January 1 of year y, the investor’s account balance was X. The investor then deposited 100 at the end of every quarter. On May 1 of year (y + 10), the account balance was 1.9X.

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

• [[math]]\frac{1.9X}{\left(1.0105\right)^{\frac{124}{3}}}+\sum_{k=1}^{42}\frac{100}{\left(1.0105\right)^{k-1}}=X [[/math]]
• [[math]]X+\sum_{k=1}^{42}{\frac{100}{(1.0035)^{3(k-1)}}}={\frac{1.9X}{(1.0035)^{124}}} [[/math]]
• [[math]]X+\sum_{k=1}^{41}\frac{100}{(1.0035)^{3k}}=\frac{1.9X}{(1.0035)^{124}}[[/math]]
• [[math]]X+\sum_{k=1}^{41}{\frac{100}{(1.0105)^{k-1}}}={\frac{1.9X}{(1.01105)^{\frac{124}{3}}}} [[/math]]
• [[math]]X+\sum_{k=1}^{42}{\frac{100}{(1.0105)^{k-1}}}={\frac{1.9X}{(1.0105)^{\frac{124}{3}}}}[[/math]]

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

Solution: C

The monthly interest rate is 0.042/12 = 0.0035. The quarterly interest rate is 1.0035³-1 = 0.0105. The investor makes 41 quarterly deposits and the ending date is 124 months from the start. Using January 1 of year y as the comparison date produces the following equation

[[math]] X+\sum_{k=1}^{41}{\frac{100}{1.0105^k}}={\frac{\textstyle1.9X}{\textstyle1.0035^{124}}} [[/math]]

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