Revision as of 22:37, 17 November 2023 by Admin (Created page with "'''Solution: E''' Let n = years. The equation to solve is <math display = "block"> \begin{array}{l}{{1000(1.03)^{2n}=2(1000)(1.0025)^{12n}}}\\ {{2n\ln1.03+\ln1000=12n\ln1.0025+\ln2000}}\\ {{m=23.775.}}\end{array} </math> This is 285.3 months. The next interest payment to Lucas is at a multiple of 6, which is 288 months. {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 17'23
Answer
Solution: E
Let n = years. The equation to solve is
[[math]]
\begin{array}{l}{{1000(1.03)^{2n}=2(1000)(1.0025)^{12n}}}\\ {{2n\ln1.03+\ln1000=12n\ln1.0025+\ln2000}}\\ {{m=23.775.}}\end{array}
[[/math]]
This is 285.3 months. The next interest payment to Lucas is at a multiple of 6, which is 288 months.
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.