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anonymous
 3 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?
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is 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
 3 years ago
Best ResponseYou've already chosen the best response.0i think it is a geometric sequence, just not sure where the ending term is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thought it was the A=R[(1+r/m)^mt1/(r/M)] there is another way, but my instructor said to figure it out? I am clueless, honestly

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think after 12 years you have \[Pr+Pr^2+Pr^3+...+Pr^{12}\] with \(P=1200\) and \(r=1.05\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we can write this as \[Pr(1+r+r^2+r^3+...+r^{11})=Pr\left(\frac{r^{12}1}{r1}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or even \[P\left(\frac{r^{13}r}{r1}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so try \[1200\left(\frac{(1.05)^{13}1.05}{.05}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the answer is here with no explanation though, maybe we can check it http://www.chegg.com/homeworkhelp/questionsandanswers/2meaganinvests1200yearira12yearsaccountearned5compoundedannuallyend12yearsq3017104

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, was trying it in the calculator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh damn i was off by an exponent thought i might have been

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try this \[1200\left(\frac{(1.05)^{12}1}{.05}\right)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this 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
 3 years ago
Best ResponseYou've already chosen the best response.0copy and past into your browser and you will see that it is the same

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i probably should have started at \[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will work out the problem with that formula. Makes sense when you see it like this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you again for your help :)
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