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First term of an infinite geometric sequence

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data
    The sum of an infinite geometric sequence is 131/2, and the sum of the first three terms is 13. Find the first term.


    2. Relevant equations
    S = a/(1-r)
    Sn = a-arn/(1-r)


    3. The attempt at a solution
    a/(1-r) = 131/2

    a-ar3/(1-r) = 13

    2a = 27-27r ........................ 1
    a-ar3 = 13-13r..... 2

    I'm stuck.
     
  2. jcsd
  3. Nov 23, 2011 #2
    [itex]2a=27-27r[/itex] (1)
    [itex]a-ar^3=13-13r[/itex] (2)

    Put 27 in evidence in (1) and put 13 in evidence in (2).

    [itex]2a=27(1-r)[/itex] (1)
    [itex]a-ar^3=13(1-r)[/itex] (2)

    Now divide (2) by (1).
     
  4. Nov 23, 2011 #3

    HallsofIvy

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    Oh!! It is the "a" that cancels allowing you to first solve for r. I was too focused on finding a. Very good pc2-brazil!
     
    Last edited: Nov 23, 2011
  5. Nov 23, 2011 #4
    I found that a =9.
     
  6. Nov 23, 2011 #5
    That is correct. I suppose you also found that r = 1/3.
    Then, the sum of the first three terms is 13:
    9 + 3 + 1 = 13
    The sum of the infinite geometric sequence is 13.5:
    9 + 3 + 1 + 1/3 + ... = [itex]\frac{9}{1-\frac{1}{3}}=13.5[/itex]
     
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