Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of interest

[[math]] \delta_{t}={\frac{t^{2}}{100}},\,t\gt0. [[/math]]

The amount of interest earned from time 3 to time 6 is also X

Calculate X.

• 385
• 485
• 585
• 685
• 785

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

Solution: E

The accumulation function is

[[math]] a(t)=\exp\biggl[\int_{0}^{t}(s^{2}/100)d s\biggr]=\exp(t^{3}/300). [[/math]]

The accumulated value of 100 at time 3 is [math]100 \exp(3^3 / 300) = 109.41743.[/math]

The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus

[[math]] \begin{array}{l}{{\left(109.41743+X\right)[a(6)/a(3)-1]=X}}\\ {{\left(109.41743+X\right)(2.0544332/1..0941743-1)=X}}\\ {{\left(109.41743+X\right)(3.877613=X}}\\ {{X=784.61.}}\end{array} [[/math]]

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