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Math Help - Geometric and Arithmetic series question

  1. #1
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    Geometric and Arithmetic series question

    The question is:

    The numbers x, y, z are succesive terms of a G.S with the sum of the three terms equal to 9. When taken in the order y, x ,z the numbers form an A.S. Find the values of x, y, z?

    G.S 9 = x(r^3-1)/(r-1)
    A.S 9 =(3/2) * (2x + (3-1)*d

    So I have that the sum of the G.S and A.S are equal and therefore I can say that the equations are equal to each other but i just don't know where to go from here or if I'm correct...

    Thanks for any help you can give me
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  2. #2
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    Quote Originally Posted by smplease View Post
    The question is:

    The numbers x, y, z are succesive terms of a G.S with the sum of the three terms equal to 9. When taken in the order y, x ,z the numbers form an A.S. Find the values of x, y, z?

    G.S 9 = x(r^3-1)/(r-1)
    A.S 9 =(3/2) * (2x + (3-1)*d

    So I have that the sum of the G.S and A.S are equal and therefore I can say that the equations are equal to each other but i just don't know where to go from here or if I'm correct...

    Thanks for any help you can give me
    "The numbers x, y, z are succesive terms of a G.S ...." => y/x = z/y => y^2 = xz .... (1)

    ".... with the sum of the three terms equal to 9" => x + y + z = 9 .... (2)

    "When taken in the order y, x ,z the numbers form an A.S." => x - y = z - x => 2x = y + z .... (3)

    Solve equations (1), (2) and (3) simultaneously for x, y, z.
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  3. #3
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    Hello, smplease!

    The numbers x, y, z are succesive terms of a G.S.
    with the sum of the three terms equal to 9.
    When taken in the order y, x ,z, the numbers form an A.S.
    Find the values of x, y, z
    The three terms are: . a,\:ar,\:ar^2

    Their sum is 9: . a + ar + ar^2\:=\:9\quad\Rightarrow\quad a(1 + r + r^2) \:=\:9\;\;{\color{blue}[1]}


    We are told: .  ar,\:a,\:ar^2 form an A.S.

    If the common difference is d, we have: . \begin{array}{cccccccc}ar + d &=& a & \Rightarrow & d &=&a(1-r) & {\color{blue}[2]}\\ a +d &=& ar^2 & \Rightarrow & d &=& a(r^2-1) & {\color{blue}[3]}\end{array}

    Eqsuate [2] and [3]: . a(1-r) \:=\:a(r^2-1)\quad\Rightarrow\quad r^2 + r - 2 \:=\:0

    . . (r-1)(r+2) \:=\:0\quad\Rightarrow\quad r \:=\:1,\:-2


    If r = 1, we have the trivial sequence: . 3,\:3,\:3


    If r = -2, substitute into [1]: . a(1 - 2 + 4) \:=\:9\quad\Rightarrow\quad a \:=\:3

    The geometric sequence is: . 3,\:-6,\:12 . . . which has a sum of 12.

    The arithmetic sequence is: . -6,\:3,\:12 . . . which has d = 9


    Therefore: . \begin{Bmatrix}x &=& 3 \\ y &=&\text{-}6 \\ z &=& 12 \end{Bmatrix}

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