Question

Geometric Sequences

Answer 1

ffbg

Answer 2

i have no idea, but off the top of my head i would write is as
\[a+ar+ar^2+ar^3+...+ar^{n-1}=\frac{a(1-r^n)}{1-r}\]

Answer 3

do u take connections?

Answer 4

you want to try it with an example? maybe that will shed light on the problem
or maybe just a sequence with 4 or 5 terms

Answer 5

i am getting somewhere, but not exactly what we want
did you add
\[\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+...+\frac{1}{ar^{n-1}}\]?

Answer 6

I'm not actually sure how I would add that with so many unknown variables...

Answer 7

common denominator is \(ar^{n-1}\)

Answer 8

adding backwards, you get
\[\frac{1+r+r^2+r^3+...+r^{n-1}}{ar^{n-1}}\]

Answer 9

oh maybe this will work nicely
\[S=a+ar+ar^2+ar^3+...+ar^{n-1}=\frac{a(1-r^n)}{1-r}\]
\[T=\frac{1+r+r^2+r^3+...+r^{n-1}}{ar^{n-1}}\]

Answer 10

hell maybe we don't even need the formula for summation
maybe we can just divide

Answer 11

\[S=a+ar+ar^2+ar^3+...+ar^{n-1}=a(1+r+r^2+r^3+...+r^{n-1})\]
\[T=\frac{1+r+r^2+r^3+...+r^{n-1}}{ar^{n-1}}\]

Answer 12

lol you get it right away!
\[\frac{S}{T}=a(1+r+r^2+r^3+...+r^{n-1})\times \frac{ar^{r-1}}{1+r+r^2+r^3+...+r^{n-1}}\]

Answer 13

leaving only \(a\times ar^{r-1}=a_1\times a_n\)

Answer 14

ohhhhhhhhhhhhhhhhh.
I get it now. Thanks.

Answer 15

yw