Revision as of 12:08, 18 November 2023 by Admin (Created page with "'''Solution: B''' The initial payment, X is <math display = "block"> \begin{aligned} 200,000 &=X\left({\frac{1}{1.03}}+{\frac{1.02}{1.03^{2}}}+\dots+{\frac{1.02^{19}}{1.03^{2}}}\right)=X{\frac{1/1.03-1.02^{20}\;/1.03^{21}}{1-1.02/1.03}}=17.7267 \\ X &= 11,282.42. \end{aligned} </math> The final payment is 11,282.42 (1.02)<sup>19</sup> =16, 436.36. {{soacopyright | 2023 }}")
Exercise
ABy Admin
Nov 18'23
Answer
Solution: B
The initial payment, X is
[[math]]
\begin{aligned}
200,000 &=X\left({\frac{1}{1.03}}+{\frac{1.02}{1.03^{2}}}+\dots+{\frac{1.02^{19}}{1.03^{2}}}\right)=X{\frac{1/1.03-1.02^{20}\;/1.03^{21}}{1-1.02/1.03}}=17.7267 \\
X &= 11,282.42.
\end{aligned}
[[/math]]
The final payment is 11,282.42 (1.02)19 =16, 436.36.
Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.