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Homework Help: Geometric progressions, i seem to be messing up on simple algebra

  1. Nov 14, 2006 #1
    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. jcsd
  3. Nov 14, 2006 #2
    [tex] 4\times 3^{n-2} = \frac{4}{3^{2-n}} = (\frac{2}{3})^{2}\times 3^{n} = \frac{4}{9}\times \;3^{n}[/tex]
     
    Last edited: Nov 14, 2006
  4. Nov 15, 2006 #3
    thanks courtrigrad!
     
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