An investor’s new savings account earns an annual effective interest rate of 3% for each of the first ten years and an annual effective interest rate of 2% for each year thereafter. The investor deposits an amount X at the beginning of each year, starting with year 1, so that the account balance just after the deposit in the beginning of year 26 is 100,000.

Determine which of the following is an equation of value that can be used to solve for X.

• [[math]]{\frac{1(00,000}{(1.03)^{10}(1.02)^{15}}}=X\sum_{k=1}^{11}{\frac{1}{(1.03)^{k-1}}}+X\sum_{k=12}^{26}{\frac{1}{(1.03)^{10}(1.02)^{k-11}}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{16}}}=X\sum_{k=1}^{10}{\frac{1}{(1.03)^{k}}}+X\sum_{k=11}^{26}{\frac{1}{(1.03)^{10}(1.02)^{k-10}}} [[/math]]
• [[math]]\frac{100,000}{\left(1.03\right)^{10}(1.02)^{16}}=X\sum_{k=1}^{11}\frac{1}{\left(1.03\right)^{k-1}}+X\sum_{k=11}^{26}\frac{1}{\left(1.03\right)^{10}(1.02)^{k-11}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{15}}}=X\sum_{k=1}^{11}{\frac{1}{(1.03)^{k-1}}}+X\sum_{k=12}^{26}{\frac{1}{(1.02)^{k-1}}} [[/math]]
• [[math]]{\frac{100,000}{(1.03)^{10}(1.02)^{16}}}=X\sum_{k=1}^{10}{\frac{1}{(1.03)^{k}}}+X\sum_{k=11}^{26}{\frac{1}{(1.02)^{k}}}[[/math]]

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

Solution: D

Let t represent the number of years since the beginning of year 1. Since the annual effective interest rate is 3% in each of years 1 through 10, and 2% each year thereafter, the present value of an amount is calculated by multiplying it by a discounting factor of

[[math]]\frac{1}{1.03^t}\,\,\mathrm{if} \, 0 \lt t \leq 10\,[[/math]]

and

[[math]]\frac{1}{(1.03)^{10}(1.02)^{t-10}}\,\,\mathrm{if} \, t \gt10\,[[/math]]

The balance is initially 0 (the account is new before the first deposit). Deposits of X are made at times t = 0, 1, 2, 3,..., 25, or equivalently at time t= k-1 for each whole number k from 1 to 26 inclusive.

For the final balance to become 0, a withdrawal of 100,000 at time t = 25 would be needed. Since the net present value of the cash flows (withdrawals minus deposits) must be zero, in a time period from a zero balance to another zero balance, we have

[[math]] \begin{aligned} {\frac{100,000}{\left(1.03\right)^{10}\left(1.02\right)^{25-10}}}-X\sum_{k=12}^{12}{\frac{1}{\left(1.03\right)^{k-1}}}-X\sum_{k=12}^{26}\frac{1}{\left(1.03\right)^{10}\left(1.02\right)^{k-1-10}}=0 \\ {\frac{100,000}{(1.03)^{10}(1.02)^{15}}}=X\sum_{k=1}^{11}{\frac{1}{(1.03)^{k-1}}}+X\sum_{k=1}^{26}{\frac{1}{(1.03)^{10}(1.02)^{k-11}}} \end{aligned} [[/math]]

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