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Math Help - Tricky One Here!

  1. #1
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    Tricky One Here!

    Hey guys, (and girls), I'm new here and I was wondering if someone could get me started on this problem:

    Find y so that (4y+1)+(y+4)+(10-y) form a geometric series.

    So far, I know the following:

    T1=4y+1
    T2=y+4
    T3=10-y

    Tn=ar^(n-1)

    Thx in advance!

    EDIT: I also know that I need to find 2 anwsers for r.
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  2. #2
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    Quote Originally Posted by Gasser666 View Post
    Hey guys, (and girls), I'm new here and I was wondering if someone could get me started on this problem:

    Find y so that (4y+1)+(y+4)+(10-y) form a geometric series.

    So far, I know the following:

    T1=4y+1
    T2=y+4
    T3=10-y

    Tn=ar^(n-1)

    Thx in advance!

    EDIT: I also know that I need to find 2 anwsers for r.
    You would know that for a geometric series r = \frac{T_{n+1}}{T_n}. In other words, the ratio of a term to the previous term is r.

    So you require r = \frac{y+4}{4y+1} = \frac{10-y}{y+4}.

    \frac{y+4}{4y+1} = \frac{10-y}{y+4} \Rightarrow (y + 4)^2 = (4y + 1)(10 - y).

    Expand, simplify and r-arrange into quadratic = 0. Solve the quadratic for y. Sub the value of y into either r = \frac{y+4}{4y+1} or r = \frac{10-y}{y+4} to get r.
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  3. #3
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    Thanks

    Thanks alot, that seems alot easier now
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