Question

Life Contingencies:

Given this life table with an interest rate of 6%

40 33 519 33 485 33 44042 41 33 467 33 428 33 378 43 42 33 407 33 365 33 309 44 43 33340 33 294 33 23 45 44 33 265 33 213 33

We are trying to solve for (IA)[40]:4 and this is the given solutionThe value of (IA)140|д is calculated in the following manner as shown below: [40141 (IA 40 401+1 [40] ( 34×(1.06))-(2x45x(1.

This is the solution, but I do not understand how they are solving for the d variables. If someone could show more work and go into detail on how they are getting the 34, 45, 62, and 69 values that would be a lot of help. Thank you!

40 33 519 33 485 33 44042 41 33 467 33 428 33 378 43 42 33 407 33 365 33 309 44 43 33340 33 294 33 23 45 44 33 265 33 213 33 143 46
The value of (IA)140|д is calculated in the following manner as shown below: [40141 (IA 40 401+1 [40] ( 34×(1.06)")-(2x45x(1.06,-)-(3x62x(1.06)-3)-(4x698(1.06)-4 ) +(4x33309x(1.06) 33519 3.16305
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Answer #1

dx = lx - lx+1

d[40] = 33519 - 33485 = 34.

d[40]+1 = 33485 - 33440 = 45

d[42] = 33407 - 33365 = 42

d[43] = 33340 - 33294 = 46

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