(i) An excerpt from a select and ultimate life table with a select period of 3 years:

[math]x[/math]	[math]l_{[x]}[/math]	[math]l_{[x]+1}[/math]	[math]l_{[x]+2}[/math]	[math]l_{x+3}[/math]	[math]x+3[/math]
60	80,000	79,000	77,000	74,000	63
61	78,000	76,000	73,000	70,000	64
62	75,000	72,000	69,000	67,000	65
63	71,000	68,000	66,000	65,000	66

(ii) Deaths follow a constant force of mortality over each year of age

Calculate [math]1000_{2 \mid 3} q_{[60]+0.75}[/math].

• 104
• 117
• 122
• 135
• 142

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

Answer: B

Under constant force over each year of age, [math]l_{x+k}=\left(l_{x}\right)^{1-k}\left(l_{x+1}\right)^{k}[/math] for [math]x[/math] an integer and [math]0 \leq k \leq 1[/math].

[[math]] \begin{aligned} & { }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}} \\ & l_{[60]+0.75}=(80,000)^{0.25}(79,000)^{0.75}=79,249 \\ & l_{[60]+2.75}=(77,000)^{0.25}(74,000)^{0.75}=74,739 \\ & l_{[60]+5.75}=(67,000)^{0.25}(65,000)^{0.75}=65,494 \end{aligned} [[/math]]

[[math]]{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679[[/math]]

[[math]]1000_{2 \mid 3} q_{[60]+0.75}=116.8[[/math]]

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