## anonymous one year ago Find the sum of a 10-term geometric sequence when the first term is 3 and the last term is 59,049 and select the correct answer below.

1. anonymous

2. pooja195

$\huge~\rm~a_n=a_1\times~r^{n-1}$ $\huge~\rm~a_n=3\times~r^{n-1}$ $\huge~\rm~a_{10}=3\times~r^9$ $\huge~\rm~59049=3\times~r^9$ $\huge~\rm~59049/3=r^9$

3. pooja195

$\huge~\rm~19683=r^9$ $\huge~\rm~\sqrt[9]{19683=r}$ From that we get r=3

4. anonymous

I got 2187? I think I did it wrong

5. pooja195

Ye you are wrong :) look at the work above

6. anonymous

I dont know how to type that into my calculator :( I originally just did 19683 times 1/9

7. pooja195

Well cacls wont be there all your life :(

8. anonymous

haha I suppose :)

9. pooja195

$\huge~\rm~\Large S_{10} = \frac{3 (1-3^{10})}{1-3} =88572$

10. anonymous

:) So I was right?

11. pooja195