Uniform annual series and future value
Suppose that there is a series of “n” uniform payments, uniform in amount and uniformly spaced, such as a payment every year. Let “A” be the amount of each uniform payment.
Let “F” be a future, single amount equivalent to the series, with “F” occurring at the same time as the last “A” payment. Then the relationship between F and A is:
F = A [ (1 + i) n – 1 ] / i
Example: If $100 is invested at the end of each year for the next 10 years in a savings account that pays 5% interest, how much will be in the account immediately after the tenth payment?
F is the unknown.
A = $100 per year
i = 5%, understood to be 5% per year, compounded annually.
n = 10 years
F = A [ (1 + 0.05) 10 – 1 ] / 0.05
= $100 [ (1.05) 10 – 1 ] / 0.05
= $100 (0.6289 / 0.05) = $1,258.
Or, using the 5% interest table, which is quicker:
F = A (F/A,5%,10) = $100 ( 12.578 ) = $1,258.
More Interest Formulas
Suppose that $1,000 is invested quarterly at 6% interest, compounded quarterly. How much will be in the account after five years?
Choose an answer by clicking on one of the letters below, or click on “Review topic” if needed.
A F = A (F/A,6%,5) = $1,000 (5.637) = $5,637
B F = A (F/A,1.5%,5) = $1,000 (5.152) = $5,152
C F = A (F/A,1.5%,20) = $1,000 (23.124) = $23,124
D F = A (A/F,1.5%,20) = $1,000 (0.0432) = $43.20
Suppose an investor plans to make monthly deposits into an account that pays 9% interest, compounded monthly, so that $100,000 will be in the account immediately after the payment at the end of Year 10. The first payment will occur at the end of Month 1 (one month from the present). How much must be deposited monthly?
Choose an answer by clicking on one of the letters below, or click on “Review topic” if needed.
A A = F (F/A,0.75%,120) = $100,000 (193.517) = $19,351,700 per month
B A = F (A/F,0.75%,120) = $100,000 (0.00517) = $517 per month
C A = F (A/F,9%,10) = $100,000 (0.0658) = $6,580 per month
D A = F (A/F,0.75%,10) = $100,000 (0.0967) = $9,670 per month
