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dawn6425
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?
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
i think it is a geometric sequence, just not sure where the ending term is
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
i think after 12 years you have \[Pr+Pr^2+Pr^3+...+Pr^{12}\] with \(P=1200\) and \(r=1.05\)
we can write this as \[Pr(1+r+r^2+r^3+...+r^{11})=Pr\left(\frac{r^{12}-1}{r-1}\right)\]
or even \[P\left(\frac{r^{13}-r}{r-1}\right)\]
so try \[1200\left(\frac{(1.05)^{13}-1.05}{.05}\right)\]
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
ok, was trying it in the calculator
oh damn i was off by an exponent thought i might have been
Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying
try this \[1200\left(\frac{(1.05)^{12}-1}{.05}\right)\]
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
copy and past into your browser and you will see that it is the same
i probably should have started at \[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer
I will work out the problem with that formula. Makes sense when you see it like this
Thank you again for your help :)