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Finding x in a geometric progression, given the sum.

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data


    [tex]1 + 2x + 4x^2 + ... = \frac{3}{4}[/tex]

    find the value of x. [Edit: Forgot to ask the question]

    2. Relevant equations

    [tex]S_n = \frac{a(1 - r^n)}{1 - r} [/tex]

    [tex] t_n = ar^{n-1} [/tex]

    3. The attempt at a solution

    a = 1

    r = 2x

    I try to solve [tex]S_n[/tex] and end up with

    [tex]2x^n = \frac{6x - 7}{4}[/tex]

    which I can't solve.

    I try to solve by equating t2 and t3 and getting x = (1/2). Which is wrong.

    Any help appreciated.
    Last edited: Sep 15, 2010
  2. jcsd
  3. Sep 14, 2010 #2
    Is the sum is to infinity?
    or to 'n terms'?

    If it is to infinity, Apply limit to your equation.
  4. Sep 14, 2010 #3


    User Avatar
    Science Advisor

    If this is an infinite sum the formula is
    [tex]S_\infty= \frac{a}{1- r}[/tex]

    If it is a finite sum, you would need to know how many terms so that "n" would be an actual integer, not a variable.

    The "..." at the end of the sum indicates it is an infinite sum.

  5. Sep 15, 2010 #4
    I've edited the original post. The problem is to find the value of x.

    Its not an infinite sum. The only solution I can see is to solve for n in the infinite series and the summation, and try to solve simultaneously.
  6. Sep 15, 2010 #5
    I had to read your answer a couple of times. I get it now. Its an infinite sum. No powers to work out.

    Thank you so much.
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