Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual effective interest rate of [math]i[/math]. The accumulated amount in the account at the end of 40 years is [math]X[/math], which is 5 times the accumulated amount in the account at the end of 20 years.

Calculate [math]X[/math]

• 4695
• 5070
• 5445
• 5820
• 6195

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

Solution: E

From basic principles, the accumulated values after 20 and 40 years are

[[math]] \begin{align*} 100[(1+i)^{20}+(1+i)^{16}+\cdots+(1+i)^{4}]=100\frac{(1+i)^{4}-(1+i)^{24}}{1-(1+i)^{4}} \\ 100[(1+i)^{40}+(1+i)^{36}+\cdots+(1+i)^{4}]=100{\frac{(1+i)^{4}-(1+i)^{44}}{1-(1+i)^{4}}} \end{align*} [[/math]]

The ratio is 5, and thus (setting [math]x=(1+i)^4[/math]))

[[math]] \begin{array}{l}{{5=\frac{(1+i)^{4}-(1+i)^{44}}{(1+i)^{4}-(1+i)^{24}}=\frac{x-x^{11}}{x-x^{6}}}}\\ {{5x-5x^{5}=x-x^{11}}}\\ {{(x^{5}-1)(x^{5}-4)=0}}\end{array} [[/math]]

Only the second root gives a positive solution. Thus

[[math]] \begin{align*}x^{5} &=4\\ x &= 1.31951 \\ x &= 1.31951 \\ X &= 100 \frac{1.31951-1.31951^{11}}{1-1.31951} = 6195.\end{align*} [[/math]]

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