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dawn6425 Group Title

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?

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    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

    • 2 years ago
  2. satellite73 Group Title
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    i think it is a geometric sequence, just not sure where the ending term is

    • 2 years ago
  3. dawn6425 Group Title
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    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

    • 2 years ago
  4. satellite73 Group Title
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    i think after 12 years you have \[Pr+Pr^2+Pr^3+...+Pr^{12}\] with \(P=1200\) and \(r=1.05\)

    • 2 years ago
  5. satellite73 Group Title
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    we can write this as \[Pr(1+r+r^2+r^3+...+r^{11})=Pr\left(\frac{r^{12}-1}{r-1}\right)\]

    • 2 years ago
  6. satellite73 Group Title
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    or even \[P\left(\frac{r^{13}-r}{r-1}\right)\]

    • 2 years ago
  7. satellite73 Group Title
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    so try \[1200\left(\frac{(1.05)^{13}-1.05}{.05}\right)\]

    • 2 years ago
  8. satellite73 Group Title
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    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

    • 2 years ago
  9. dawn6425 Group Title
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    ok, was trying it in the calculator

    • 2 years ago
  10. satellite73 Group Title
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    oh damn i was off by an exponent thought i might have been

    • 2 years ago
  11. dawn6425 Group Title
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    Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying

    • 2 years ago
  12. satellite73 Group Title
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    try this \[1200\left(\frac{(1.05)^{12}-1}{.05}\right)\]

    • 2 years ago
  13. satellite73 Group Title
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    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

    • 2 years ago
  14. satellite73 Group Title
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    copy and past into your browser and you will see that it is the same

    • 2 years ago
  15. satellite73 Group Title
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    i probably should have started at \[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer

    • 2 years ago
  16. dawn6425 Group Title
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    I will work out the problem with that formula. Makes sense when you see it like this

    • 2 years ago
  17. dawn6425 Group Title
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    Thank you again for your help :)

    • 2 years ago
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