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dawn6425
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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
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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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i think it is a geometric sequence, just not sure where the ending term is
 2 years ago

dawn6425 Group TitleBest ResponseYou've already chosen the best response.0
I 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i think after 12 years you have \[Pr+Pr^2+Pr^3+...+Pr^{12}\] with \(P=1200\) and \(r=1.05\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
we can write this as \[Pr(1+r+r^2+r^3+...+r^{11})=Pr\left(\frac{r^{12}1}{r1}\right)\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
or even \[P\left(\frac{r^{13}r}{r1}\right)\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
so try \[1200\left(\frac{(1.05)^{13}1.05}{.05}\right)\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the answer is here with no explanation though, maybe we can check it http://www.chegg.com/homeworkhelp/questionsandanswers/2meaganinvests1200yearira12yearsaccountearned5compoundedannuallyend12yearsq3017104
 2 years ago

dawn6425 Group TitleBest ResponseYou've already chosen the best response.0
ok, was trying it in the calculator
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oh damn i was off by an exponent thought i might have been
 2 years ago

dawn6425 Group TitleBest ResponseYou've already chosen the best response.0
Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
try this \[1200\left(\frac{(1.05)^{12}1}{.05}\right)\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
copy and past into your browser and you will see that it is the same
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i probably should have started at \[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer
 2 years ago

dawn6425 Group TitleBest ResponseYou've already chosen the best response.0
I will work out the problem with that formula. Makes sense when you see it like this
 2 years ago

dawn6425 Group TitleBest ResponseYou've already chosen the best response.0
Thank you again for your help :)
 2 years ago
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