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Math Help - [SOLVED] Sequences (geometric)

  1. #1
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    Exclamation [SOLVED] Sequences (geometric)

    x, y and z are consecutive terms in a geometric sequence. If
    x+y+z =7/3, and
    x^2 + y^2 + z^2 = 91/9, find the values of x, y and z.
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  2. #2
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    Quote Originally Posted by HelenaStage View Post
    x, y and z are consecutive terms in a geometric sequence. If
    x+y+z =7/3, and
    x^2 + y^2 + z^2 = 91/9, find the values of x, y and z.
    The general form of a geometric series is \sum_{k=1}^n ar^k

    So x=x, and y in the next term so y=rx and z=ry=r^2x

    So x+rx+r^2x=\frac{7}{3}=x(1+r+r^2)

    And the second equation tells us

     <br />
x^2+(rx)^2+(r^2x)^2=\frac{91}{9}=x^2+r^2x^2+r^4x^2  =x^2(1+r^2+r^4)<br />

    So you just have to solve those equations
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  3. #3
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    Quote Originally Posted by artvandalay11 View Post
    The general form of a geometric series is \sum_{k=1}^n ar^k

    So x=x, and y in the next term so y=rx and z=ry=r^2x

    So x+rx+r^2x=\frac{7}{3}=x(1+r+r^2)

    And the second equation tells us

     <br />
x^2+(rx)^2+(r^2x)^2=\frac{91}{9}=x^2+r^2x^2+r^4x^2  =x^2(1+r^2+r^4)<br />

    So you just have to solve those equations
    How? By substitution I get a quite complicated equation with radicals...
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  4. #4
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    it seems to me that the best way to go about it is substitute for x, and yes it is going to get messy

    using an equation solver it looks like x=1/3 and r=-3 or x=3 and r=-1/3

    so those are the solutions you should get
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  5. #5
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    Quote Originally Posted by HelenaStage View Post
    How? By substitution I get a quite complicated equation with radicals...
    HI

    You might not like this

    x(1+r+r^2)=\frac{7}{3} --- 1

    x^2(1+r^2+r^4)=\frac{91}{9} --- 2

    From 1

    [x(1+r+r^2)]^2=[\frac{7}{3}]^2

    x^2(1+2r+3r^2+2r^3+r^4)=\frac{49}{9} --- 3

    so now 3/2

     <br />
\frac{x^2(1+2r+3r^2+2r^3+r^4)}{x^2(1+r^2+r^4)}=\fr  ac{49}{9}\times\frac{9}{91}<br />

    21r^4+91r^3+112r^2+91r+21=0

    Since its a quartic ,

     <br /> <br />
21(r^2+\frac{1}{r^2})+91(r+\frac{1}{r})+112=0<br />

    Let r+\frac{1}{r}=p

    21(p^2-2)+91p+112=0\Rightarrow p=-1(ommitted) , - \frac{10}{3}

    therefore p=-\frac{10}{3}

    r+\frac{1}{r}=-\frac{10}{3}

    r=-3 , -1/3

    so now head back to find x .
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  6. #6
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    Oh, i substituted before squaring...^^
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