Exercise
ABy Admin
Nov 18'23
Answer
Solution: D
[[math]]
\overline{{{a}}}_{\overline{{{20}}|}}=1.5\overline{{{a}}}_{\overline{{{10}}|}},\;\;\;\frac{1-e^{-20\delta}}{\delta}=1.5\frac{1-e^{-10\delta}}{\delta},\;\;\;e^{-20\delta}-1.5e^{-10\delta}+0.5=0.[[/math]]
Let [math]X = e^{-10\delta}[/math]. We then have the quadratic equation
[[math]]
X^2 -1.5X + 0.5 = 0
[[/math]]
with solution [math]X = 0.5[/math] for [math]\delta = \ln0.5\,/\,(-10)=0.069315. [/math]
Then, the accumulated value of a 7-year continuous annuity of 1 is
[[math]]
\overline{{{s}}}_{\overline{{{7}}}|}=\frac{e^{7(0.069315)}-1}{0.069315}=9.01\,.
[[/math]]