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Geometric progression

  1. Nov 7, 2006 #1
    ok, the qn goes like this..

    Find the first term and common ration of a geometric progression if the sum to n is given by
    6 - 2/(3^(n-1))

    I tried solving by making both terms of the eqn having the same denominator by multiplying (3^(n-1)) to the first term 6 and then by taking out the 2, i am able to make the formula to 2(3^n - 1)/ (3^n-1). But what about the denominator, i remember that there must be no n at the bottom. Only r - 1 where in this case, r is to be found..

    Anyone out here can give me some guide? Thanks for any help given.
  2. jcsd
  3. Nov 8, 2006 #2


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    Science Advisor

    First, you have written that wrong. The denominator is not 3^n- 1, it is 3^(n-1). You should have
    [tex]\frac{6(3^{n-1})- 2}{3^{n-1}}= 2(3^n- 1}{3^{n-1}[/tex]

    But it's really not necessary to use a general formula. Since you only need to find two numbers, a and r for the general term arn, you really only need two equations. The first term, with n= 1, is a. The formula you are given says it must be
    [tex]a= 6- \frac{2}{3^{1-1}}[/tex]
    The sum of the first two terms, with n= 2 is a+ ar and so
    [tex]a+ ar= 6- \frac{2}{3^{2-1}}[/tex]
    Can you solve those two equations for a and r?
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