## 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?

1. satellite73

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

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

3. dawn6425

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

i think after 12 years you have $Pr+Pr^2+Pr^3+...+Pr^{12}$ with $$P=1200$$ and $$r=1.05$$

5. satellite73

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

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

7. satellite73

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

8. satellite73

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

ok, was trying it in the calculator

10. satellite73

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

11. dawn6425

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

12. satellite73

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

13. satellite73

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

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

15. satellite73

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

16. dawn6425

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

17. dawn6425

Thank you again for your help :)