# Geometric progressions, i seem to be messing up on simple algebra

1. Nov 14, 2006

### mr_coffee

hello everyone, i'm trying to figure out this geometric progession and i'm checking it with my calculator and i'm definatley not getting what they are getting:

I'm trying to find the term after the last term, I found the ratio which is 3/2, now I must multiply that ratio by the last term which is:
(2^3)*3^(n-3)
So i'd have:
(3/2)*(2^3)*3^(n-3)
The calculator spits out:
(4/9)*3^n

I did this problem before and I got that I dont know where my brain is, but now i'm getting:

4*3^(n-2);

Becuase the 2 will cancel out the 8, and leave you 4, and 3 x 3^(n-3) is like 3^(n-3+1) = 3^(n-2)

What am I missing?

Here is the geometric progression and below is my work, once I apply the formula its simple aglebra and adding of exponents but I keep screwing it up.
http://suprfile.com/src/1/4ds5zep/lastscan.jpg [Broken]

Thanks!

Last edited by a moderator: May 2, 2017
2. Nov 14, 2006

$$4\times 3^{n-2} = \frac{4}{3^{2-n}} = (\frac{2}{3})^{2}\times 3^{n} = \frac{4}{9}\times \;3^{n}$$