Question

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?

Answer 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

Answer 2

i think it is a geometric sequence, just not sure where the ending term is

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

or even
\[P\left(\frac{r^{13}-r}{r-1}\right)\]

Answer 7

so try
\[1200\left(\frac{(1.05)^{13}-1.05}{.05}\right)\]

Answer 8

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

Answer 9

ok, was trying it in the calculator

Answer 10

oh damn i was off by an exponent
thought i might have been

Answer 11

Well that wasnt too hard. Had no clue how to set it up. Thank you very much for replying

Answer 12

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

Answer 13

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

Answer 14

copy and past into your browser and you will see that it is the same

Answer 15

i probably should have started at
\[P+Pr+Pr^2+Pr^3+...+Pr^{11}\] and that would give the above answer

Answer 16

I will work out the problem with that formula. Makes sense when you see it like this

Answer 17

Thank you again for your help :)