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September 30, 2016
Posted by **Cindy** on Thursday, November 12, 2009 at 1:32pm.

- Math -
**tchrwill**, Thursday, November 12, 2009 at 3:21pmThe formula for determining the accumulation of a series of periodic deposits, made at the end of each period, over a given time span, an ordinary annuity, is

.......S(n) = R[(1 + i)^n - 1]/i

where S(n) = the accumulation over the period of n intervals, R = the periodic deposit, n = the number of interest paying periods, and i = the annual interest % divided by 100 divided by the number of interest paying periods per year.

When an annuity is cumputed on the basis of the payments being made at the beginning of each period, an annuity due, the total accumulation is based on one more period minus the last payment. Thus, the total accumulation becomes

S(n+1) = R[(1+i)^(n+1) - 1]/i - R

.......= R[[{(1+i)^(n+1)- 1}/i]-1]

Here, R = $450, i = (10/100)/4 = .208333 and n = 9x4 = 36. - Math -
**Cindy**, Thursday, November 12, 2009 at 3:39pmThanks tchrwill, But I'm still confused. The question has a multiply choice answer, and it's one of these:

$26,430.20

$29,073.31

$26,430.25

$27,751.79