Janet receives a $ 10,000 life insurance benefit. If she uses the proceeds to buy an n-year annuity immediate, the annual payout will be 1613.36. If a 2n-year annuity due is purchased, the annual payout will be 1507.44. Both calculations are based on an effective annual interest rate of i. Calculate i.
Having trouble with this problem. My prof told me to use annuity immediate for both of the options but I cannot seem to get the correct answer.
My approach is $k_1$ $a_n*i$=$k_2$ $a_2n*i$ and solve for i, where $k_1$=1613.26 and $k_2=1507.44$.
Not a finance student here, but let me give it a shot.First assuming that Janet will invest all of her $10,000$ dollars in assets, and the reasoning that the sum of all annual payments should be equal to the accumulated value, we have: $$(1 + i)^n * A = \sum R$$ Taking $i$ as the interest rate, $n$ as the number of periods, $A$ as the initial investment and $R$ as the periodic return. Considering we have $n$ equal periodic returns, $\sum R = n*R$. If we divide by A on both sides we have: $$(1 + i)^n = \dfrac{n*R}{A}$$ For another asset with the same interest rate $i$, initial investment $A$, but diferent duration $2n$ and periodic return $R'$, we have the same relation: $$(1 + i)^{2n} = \dfrac{2n*R'}{A}$$ But $(1 + i)^{2n} = [(1 + i)^n]^2 = (\dfrac{n*R}{A})^2 = \dfrac{n^2 * R^2}{A^2}$!
Now, it's just a question of fiddling around with this result to find the value of n and using the logarithm function to find the value of i.