Payments are made to an account at a continuous rate of [math]8k + tk[/math], where [math]0 \leq t \leq 10.[/math]

Interest is credited at a force of interest [math] \delta_t = \frac{1}{8 + t}.[/math]

After time 10, the account is worth 20,000.

Calculate k.

• 111
• 116
• 121
• 126
• 131

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

Solution: A

The accumulation function is:

[[math]] a(t)=e^{\int_{0}^{t}{\frac{1}{8+r}}d r}=e^{\ln(8+r)_{0}^{\iota}}={\frac{8+t}{8}}. [[/math]]

Using the equation of value at end of 10 years:

[[math]] \begin{align*} 20,000=\textstyle{\int_{0}^{10}}(8k+t k){\frac{a(10)}{a(t)}}d t=k\int_{0}^{10}(8+t){\frac{18/8}{(8+t)/8}}d t=k\int_{0}^{10}18d t \\ ={180k\Rightarrow k}={\frac{20,000}{180}}=111. \end{align*} [[/math]]

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