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Sum to Infinity of a Geometric Series

  1. Aug 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Q.: The numbers [itex]\frac{1}{t}[/itex], [itex]\frac{1}{t - 1}[/itex], [itex]\frac{1}{t + 2}[/itex] are the first, second and third terms of a geometric sequence.
    Find (i) the value of t,
    (ii) the sum to infinity of the series.

    2. Relevant equations

    S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

    3. The attempt at a solution

    I have already solved (i), the value of t = [itex]\frac{1}{4}[/itex].

    Ans.: From text book = 6

    Attempt at (ii): S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex]

    a = [itex]\frac{1}{t}[/itex] = [itex]\frac{1}{1/4}[/itex] = 4

    r = [itex]\frac{U2}{U1}[/itex] = [itex]\frac{1}{t-1}[/itex]/ 4

    [itex]\frac{1}{1/4 - 1}[/itex]/ 4

    [itex]\frac{1}{-3/4}[/itex]/ 4

    [itex]\frac{-4/3}{4}[/itex]

    [itex]\frac{-4}{3}[/itex]([itex]\frac{1}{4}[/itex]) = [itex]\frac{-4}{12}[/itex] = [itex]\frac{-1}{3}[/itex]

    Lastly,
    S[itex]\infty[/itex] = [itex]\frac{a}{1 - r}[/itex] = [itex]\frac{4}{1-(-1/3)}[/itex]

    [itex]\frac{4}{1 + 1/3}[/itex]

    [itex]\frac{4}{4/3}[/itex] = 4([itex]\frac{3}{4}[/itex]) = [itex]\frac{12}{4}[/itex] = 3

    I have shown this problem on another site, and the other users seem to think that the book has the answer incorrect; with 3 being the correct value. I just wanted to post my attempt here too, to get a second opinion. Thank you.
     
    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 20, 2011 #2

    micromass

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    Yes, 3 seems to be correct here.
     
  4. Aug 20, 2011 #3
    Great. Thanks for confirming that with me.
     
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