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.

Question

anonymous · Answer

177,147
 88,572
 88,575<---my answer
 177,144

pooja195 · Answer

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

pooja195 · Answer

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

anonymous · Answer

I got 2187? I think I did it wrong

pooja195 · Answer

Ye you are wrong :) look at the work above

anonymous · Answer

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

pooja195 · Answer

Well cacls wont be there all your life :(

anonymous · Answer

haha I suppose :)

pooja195 · Answer

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

anonymous · Answer

:) So I was right?

pooja195 · Answer

You were  off a few numbers look at your answer choices