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    cant seem to answer this question:

    in a geometric progression, the second term is -4 and the sum to infinity is 9, find the common ratio?

    i can see that a= -4/r and sum to infinity = a/1-r

    the answer is -1/3 (teacher gave us the answers) but i cant get to that. any help would be appreciated
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    a = -4/r is correct

    You also know the sum to infinity formula is a/1-r and you're told in the question that this equals 9

    So you know a/(1-r) = 9

    You have two equations and two unknowns, so what do you do?
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    What answer do you get? You have enough information to do the question; substitute a=-\frac{4}{r} into 9=\frac{a}{1-r} and solve for r (it's probably easiest to multiply through by 1-r first).
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    How do you know r= -1/3 as opposed to 4/3?
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    (Original post by xHuStLeR23)
    How do you know r= -1/3 as opposed to 4/3?
    Because we need -1 < r < 1 for the sum to infinity to converge.
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    Thanks for the help guys, had a bit of a blank and couldnt see that it was a quadratic giving r=4/3 and r=-1/3. i have no idea why it cant be 4/3, maybe my teacher made a mistake
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    (Original post by matt730)
    i have no idea why it cant be 4/3
    Work out the 10th, 100th and 1000th terms when it's -1/3, and again when it's 4/3, and you'll see why.
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    Work out the 10th, 100th and 1000th terms when it's -1/3, and again when it's 4/3, and you'll see why.
    ahhh now it makes sense. Its been a long day of c2 revision lol
 
 
 
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Updated: April 14, 2011
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