You are given:

(i) [math]\quad A_{35}=0.188[/math]

(ii) [math]A_{65}=0.498[/math]

(iii) [math]{ }_{30} p_{35}=0.883[/math]

(iv) [math]\quad i=0.04[/math]

Calculate [math]1000 \ddot{a}_{35: 30}^{(2)}[/math] using the two-term Woolhouse approximation.

• 17,060
• 17,310
• 17,380
• 17,490
• 17,530

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Answer: C

[[math]] \begin{aligned} \ddot{a}_{35: 30}^{(2)} & \approx \ddot{a}_{35: \overline{30}}-\frac{(m-1)}{2 m}\left(1-v^{30}{ }_{30} p_{35}\right) \\ \ddot{a}_{35: \overline{30}} & =\frac{1-A_{35: 30}}{d}=\frac{1-A_{35: 30}^{1}-{ }_{30} E_{35}}{d} \\ & =\frac{1-\left(A_{35}-{ }_{30} E_{35} \times A_{65}\right)-{ }_{30} E_{35}}{d} \end{aligned} [[/math]]

Since [math]{ }_{30} E_{35}=v^{30}{ }_{30} p_{35}=0.2722[/math], then

[[math]] \begin{aligned} & \ddot{a}_{35: 30}=\frac{1-\left(A_{35}-v^{30}{ }_{30} p_{35} \times A_{65}\right)-v^{30}{ }_{30} p_{35}}{d} \\ &=\frac{1-(0.188-(0.2722)(0.498))-0.2722}{(0.04 / 1.04)} \\ &=17.5592 \\ & \ddot{a}_{35: 30 \mid}^{(2)} \approx 17.5592-\frac{1}{4}(1-0.2722)=17.38 \\ & 1000 \ddot{a}_{35: 30}^{(2)} \approx 1000 \times 17.38=17,380 \end{aligned} [[/math]]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.