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Geometric progression related problem

  1. Sep 4, 2013 #1
    1. The problem statement, all variables and given/known data
    The sum of the squares of three distinct real numbers, which are in geometric progression is ##S^2##. If their sum is ##\alpha S##, show that ##\alpha^2 \in (1/3,1) \cup (1,3)##.


    2. Relevant equations



    3. The attempt at a solution
    Let the three numbers be ##a , \, ar, \,ar^2##. According to the question
    $$a^2+a^2r^2+a^2r^4=S^2 \, (*)$$
    and
    $$a+ar+ar^2=\alpha S \, (**)$$
    Squaring (**), I get
    $$a^2+a^2r^2+a^2r^4+2a^2r+2a^2r^3+2a^2r^2=\alpha^2S^2$$
    From (*)
    $$S^2+2a^2r(1+r+r^2)=\alpha^2S^2$$
    The equation (**) can rewritten as ##a(1+r+r^2)=\alpha S##. Using this
    $$S^2+2ar(\alpha S)=\alpha^2S^2 \Rightarrow S+2a\alpha r=\alpha^2S$$
    I am stuck here.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Sep 4, 2013 #2

    vela

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    You have the two equations
    \begin{align*}
    \alpha S &= a(1+r+r^2) \\
    S^2 &= a^2(1+r^2+r^4)
    \end{align*} Can you see how you might solve for ##\alpha^2## while eliminating ##S## and ##a## at the same time?
     
  4. Sep 5, 2013 #3
    After squaring eq (**) don't open the expression.Instead divide it by eq (*).This will eliminate 'S' and 'a' simultaneously.

    Now,the tricky part .You can write
    1+r2+r4 = (1+r+r2)(1-r+r2).

    This will eliminate the expression (1+r+r2) from Nr and Dr.

    You are now left with a quadratic equation in r2.Now in order to have a real value of 'r' the condition is ? I am sure you know this :smile:
     
    Last edited: Sep 5, 2013
  5. Sep 5, 2013 #4
    I get the following quadratic:
    $$r^2(\alpha^2-1)-r(\alpha^2+1)+\alpha^2-1=0$$
    I know I have to deal with discriminant here but do I have to use D>=0 or D>0 (D is discriminant)? I have tried using both of them and D>0 gives the correct answer but why not D>=0? :confused:
     
  6. Sep 5, 2013 #5

    vela

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    You're supposed to be solving for ##\alpha^2##, so you actually have a linear equation on your hands there. [strike]But how'd you get that? It's not correct.[/strike] Tanya basically told you how to do the whole problem.
     
    Last edited: Sep 5, 2013
  7. Sep 5, 2013 #6
    Yes I know I have to solve for ##\alpha^2## but Tanya also said that I would end up with quadratic in r. Is what I have written incorrect? From the quadratic, ##\alpha^2 \neq 1##, the other values of ##\alpha^2## can be found out using the fact that D is greater than zero for the quadratic. My doubt is why not use D>=0? D>=0 gives 3 and 1/3 as possible solutions for ##\alpha^2## while the given answer does not include them. How am I supposed to know if I have to use D>=0 or D>0? I hope you understand my query.
     
  8. Sep 5, 2013 #7

    vela

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    Sorry, your previous result was right. (I made the algebra mistake.)

    You actually don't get a quadratic; you should get a rational expression. Try plotting your result for ##\alpha^2## as a function of r.
     
  9. Sep 5, 2013 #8
    ##\alpha^2=\frac{1+r+r^2}{1-r+r^2}##

    I plotted this on Wolfram|Alpha: http://www.wolframalpha.com/input/?i=y=(1+x+x^2)/(1-x+x^2)
    I have replaced ##\alpha^2## with y and r with x. It gives the result that range of y or ##\alpha^2## is [1/3,3] which is incorrect.

    What's wrong with my method of evaluating the discriminant? I don't think the given problem requires me to do curve-sketching.
     
  10. Sep 5, 2013 #9

    vela

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    You have to consider the allowed values of r. Not every value of r will satisfy the requirements stated in the first sentence of the problem. That will rule both of the D=0 roots.

    From the graph, it's easy to see why two values of ##\alpha^2## should be excluded.
     
  11. Sep 5, 2013 #10
    I think I understand it. r cannot be equal to 0,1 and -1 because for these values, the three terms won't be distinct. So 1,3 and 1/3 are excluded.

    Thanks a lot vela and Tanya! :)
     
  12. Sep 5, 2013 #11

    vela

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    You got it.
     
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