Find the sum of a 9-term geometric sequence when the first term is 4 and the last term is 1,024.

Question

anonymous · Answer

@mathmate

anonymous · Answer

@Preetha

mathmate · Answer

Here's an article to familiarize yourself with geometric sequences and sums.
If you encounter problems while solving, please post again.

Hint:
in a geometric sequence, $ar^0=a$ is the first term, r is the common ratio, and
$ar^{n-1}$ is the nth term.
So the nineth term divided by the first term equals
$\Large \frac{ar^{9-1}}{ar^{0}}=r^8$
This will allow you to solve for r.

anonymous · Answer

why thank you, let me try it out and I'll ask you for help if needed c:

mathmate · Answer

Here's the link that I forgot to post: http://www.mathsisfun.com/algebra/sequences-sums-geometric.html

anonymous · Answer

alright so I have a 113.77778 added to the last term but it comes out to 1027 in the ninth term. Could I get some guidance here?

mathmate · Answer

If you solve r^8=1024/4, you should find an integer value for r (whole number).
The rest will work well from there.

anonymous · Answer

that means r = 2, but that makes the 8th term 1,024

anonymous · Answer

things got all out of wack, sorry, I should have tried counting better, it was r = 2 the whole time. Thank you for your time.