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dawn6425

  • 2 years ago

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?

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

  3. dawn6425
    • 2 years ago
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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

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

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

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

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

  8. satellite73
    • 2 years ago
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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

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

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

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

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

  13. satellite73
    • 2 years ago
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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

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

  15. satellite73
    • 2 years ago
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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

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

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

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