anonymous
  • anonymous
This is a rather long question, but I would really appreciate any help. Megan invests $1200 each year in an IRA for 12 years in an account that earned 5% compounded annually. At the end of 12 years, she stopped making payments to the account, but continued to invest her accumulated amount at 5% compounded annually for the next 11 years. a. What was the value of the IRA at the end of 12 years? b. What was the value of the investment at the end of the next 11 years? c. How much interest did she earn?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is there any chance you have a formula for the fist one i can grind it out, but i am sure there is an easier way
anonymous
  • anonymous
i think it is a geometric sequence, just not sure where the ending term is
anonymous
  • anonymous
I thought it was the A=R[(1+r/m)^mt-1/(r/M)] there is another way, but my instructor said to figure it out? I am clueless, honestly

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More answers

anonymous
  • anonymous
i think after 12 years you have \[Pr+Pr^2+Pr^3+...+Pr^{12}\] with \(P=1200\) and \(r=1.05\)
anonymous
  • anonymous
we can write this as \[Pr(1+r+r^2+r^3+...+r^{11})=Pr\left(\frac{r^{12}-1}{r-1}\right)\]
anonymous
  • anonymous
or even \[P\left(\frac{r^{13}-r}{r-1}\right)\]
anonymous
  • anonymous
so try \[1200\left(\frac{(1.05)^{13}-1.05}{.05}\right)\]
anonymous
  • anonymous
the answer is here with no explanation though, maybe we can check it http://www.chegg.com/homework-help/questions-and-answers/2-meagan-invests-1-200-year-ira-12-years-account-earned-5-compounded-annually-end-12-years-q3017104
anonymous
  • anonymous
ok, was trying it in the calculator
anonymous
  • anonymous
oh damn i was off by an exponent thought i might have been
anonymous
  • anonymous
Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying
anonymous
  • anonymous
try this \[1200\left(\frac{(1.05)^{12}-1}{.05}\right)\]
anonymous
  • anonymous
this one agrees with the no explanation chubb answer here is the calculation http://www.wolframalpha.com/input/?i=1200\left%28\frac{%281.05%29^{12}-1}{.05}\right%29
anonymous
  • anonymous
copy and past into your browser and you will see that it is the same
anonymous
  • anonymous
i probably should have started at \[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer
anonymous
  • anonymous
I will work out the problem with that formula. Makes sense when you see it like this
anonymous
  • anonymous
Thank you again for your help :)

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