An investor opens a savings account with no initial deposit that pays a constant 4.5% annual force of interest. The investor deposits 700 at the end of every six-month period for 20 years. The account balance immediately after the last deposit is X.

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

• [[math]]\quad \sum_{n=0}^{39} 700 e^{0.0225 n}=X e^{-0.9}[[/math]]
• [[math]]\quad \sum_{n=0}^{39} 700 e^{-0.0225 n}=X e^{0.9}[[/math]]
• [[math]]\quad \sum_{n=0}^{39} 700 e^{-0.0225 n}=X e^{-0.9}[[/math]]
• [[math]]\quad \sum_{n=0}^{39} 700 e^{0.0225 n}=X e^{0.9}[[/math]]
• [[math]]\quad \sum_{n=1}^{40} 700 e^{-0.0225 n}=X e^{-0.9}[[/math]]

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

Solution: E

If accumulating the payments, they accumulate for 0, 1, ..., 39 periods. This accumulation is reflected in the left-hand sides of (A) and (D). Since the accumulated value is X, neither right-hand side is correct. If discounting the payments to time zero, the payments are discounted for 1, 2, ..., 40 periods as reflected in the left-hand side of (E). The right-hand side of (E) discounts the accumulated value of X for 40 periods and hence is the correct value. Answers (B) and (C) do not reflect the time-zero present value.

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