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Thread: Arithmetic and geometric sequences.

  1. #1
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    Arithmetic and geometric sequences.

    (1) The lengths of the sides of a right triangle are three consecutive terms of an increasing arithmetic sequence. Show that the lengths are in the ratio 3:4:5.

    (2) If the pth and qth terms of an arithmetic sequence are q and p respectively, find the (p+q)th term.

    (3) Find X and y if the sequence 2y, 2xy, x, xy/2 .... is geometric.

    Guys i really need you help
    Thank you
    Last edited by mr fantastic; May 15th 2010 at 03:18 PM. Reason: Re-titled.
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  2. #2
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    Hello, vvc531!

    (1) The lengths of the sides of a right triangle are three consecutive terms of
    an increasing arithmetic sequence. Show that the lengths are in the ratio 3:4:5.

    The first three terms of an arithmetic sequence are: .$\displaystyle a,\:a+d,\:a+2d$

    If they are the sides of a righ triangle: .$\displaystyle a^2 + (a+d)^2 \:=\:(a+2d)^2$

    . . which simplifies to: .$\displaystyle a^2 - 2ad - 3d^2 \:=\:0$

    . . which factors: .$\displaystyle (a -3d)(a +d) \:=\:0$

    . . and has the positive root: .$\displaystyle a \:=\:3d $


    Hence, the three sides are: .$\displaystyle \begin{Bmatrix}a &=& 3d \\a+d &=& 4d \\ a+2d &=&5d \end{Bmatrix}$


    And their ratio is: .$\displaystyle 3d:4d:5d \;=\;3:4:5$




    (3) Find $\displaystyle x$ and $\displaystyle y$ if the sequence $\displaystyle 2y, 2xy, x, \tfrac{xy}{2}$ is geometric.

    The first four terms of a geomeric sequence are: .$\displaystyle a,\:ar,\:ar^2,\:ar^3$

    So we have: .$\displaystyle \begin{array}{cccc}a &=& 2y & [1] \\ ar &=& 2xy & [2] \\ ar^2 &=& x & [3] \\ ar^3 &=& \frac{xy}{2} & [4] \end{array}$


    $\displaystyle \begin{array}{ccccccc}\text{Divide }[2] \div [1] & \dfrac{ar}{a} \:=\:\dfrac{2xy}{2y} & \Rightarrow & r \:=\:x & [5] \\ \\[-4mm]
    \text{Divide }[3] \div [2] & \dfrac{ar^2}{ar} \:=\:\dfrac{x}{2xy} & \Rightarrow & r \:=\:\dfrac{1}{2y} & [6] \\ \\[-4mm]
    \text{Divide }[4] \div[3] & \dfrac{ar^3}{ar^2} \:=\:\dfrac{\frac{xy}{2}}{x} & \Rightarrow & r \:=\:\dfrac{y}{2} & [7]\end{array}$


    Equate [7] and [6]: .$\displaystyle \frac{y}{2}\:=\:\frac{1}{2y}\quad\Rightarrow\quad y^2 \:=\:1\quad\Rightarrow\quad y \:=\:\pm1$

    Substiute into [7]: .$\displaystyle r \:=\:\pm\frac{1}{2}$

    Substitute into [5]: .$\displaystyle x \:=\:\pm\frac{1}{2}$


    Therefore: .$\displaystyle (x,y) \;=\;\left(\tfrac{1}{2},\tfrac{1}{2}\right),\;\lef t(\text{-}\tfrac{1}{2},\text{-}\tfrac{1}{2}\right) $

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  3. #3
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    Thank you so much

    Thank you so much
    can you please take a look at my other post
    really hard pre-cal questions (4) ~(5) ?
    You are great!
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