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March 5, 2015

March 5, 2015

Posted by **Thara!** on Wednesday, April 21, 2010 at 8:57pm.

I'm not sure how to get the answer for this question which is:

$8649.11

- math! -
**tchrwill**, Friday, April 23, 2010 at 1:21pmSn = D[(1+i)^n-1)/i where

Sn = the accumulated sum over n interest periods

D = the periodic deposit

i = the periodic interest paid = I%/100n

n = the number of interest bearing periods

The total interest beariing period is 15 + 12 months

The monthly interest is 6/100(12) = .005

The monthly deposit is $300

Therefore,

S(27)=300[(1.005)^27-1]/.005= $8,649.11

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