Question 252920: At the end of each three months Laura puts $200 into an account which pays 10% compounded quarterly. After 10 years she discontinues the payments but leaves the total amount in the account to collect interest for 2 more years. Determine the balance in the account at the end of 12 years.
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! What we have is a geometric sequence + I = p(1+r/n)^(nt).
Think of it this way you put $200 into the account at 10% compounded quarterly for 10 years. This $200 will compound a total of 40 times. Then another $200 into the account at 10% compounded quarterly for 10 years. This second $200 will compound a total of 39 times. SO, we continue until you put your last 200 in at 10% compounded quarterly for only 0 compound periods.
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It appears that we have:
200(1+.10/4)^(40) + 200(1+.10/4)^(39) + 200(1+.10/4)^(38) + 200(1+.10/4)^(37) and so on . . . + 200(1+.10/4)^(0).
This is a geometric sum which when calculated out will be: ~$14,017.52.
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We are not done yet. This amount will now sit in the account untouched for two more years, or 8 compound periods. The formula for this is
A = 14017.72(1 + .10/4)^(4*2)
A = $17,078.99.
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I hope that helps and makes sense.
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