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Progressions and series prove this

  1. Nov 22, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If [itex]x_1,x_2,x_3.......x_n[/itex] are in H.P. then prove that [itex]x_1x_2+x_2x_3+x_3x_4...........+x_{n-1}x_n=(n-1)x_1x_n[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Since [itex]x_1,x_2,x_3.......x_n[/itex] are in H.P. therefore
    [itex] \frac{1}{x_1},\frac{1}{x_2},\frac{1}{x_3}..........,\frac{1}{x_n}[/itex] will be in A.P. Now common difference of this A.P.
    [itex]d=\dfrac{\frac{1}{x_n}-\frac{1}{x_1}}{n-1} \\
    x_1x_n=\dfrac{x_1-x_n}{d(n-1)}\\
    (n-1)x_1x_n=\dfrac{x_1-x_n}{d} [/itex]

    Looks like I've arrived at the R.H.S. But what about LHS?
     
  2. jcsd
  3. Nov 22, 2012 #2
    This the second one of your problems I've worked on. Just so you know, you need to write out your acronyms. No one has ANY clue what you mean if you don't do it. I was tempted to write something about Hewlett-Packard. :tongue2:

    Alright. [itex]x_1x_2+x_2x_3+x_3x_4+...+x_{n-1}x_{n}[/itex] is the same thing as saying [itex]x_{1}(\frac{x_{1}}{1+d}) + (\frac{x_{1}}{1+d})(\frac{x_{1}}{1+2d})+...+(\frac{x_{1}}{1+(n-2)})(\frac{x_{1}}{1+(n-1)})[/itex].

    You can probably figure it out from here. :wink:
     
  4. Nov 22, 2012 #3

    utkarshakash

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    Hey, do you see my title. It says 'Progressions and series prove this'? And I don't understand how can you relate Hewlett-Packard with progressions and series?:confused:It's not even a mathematical term. OK I will do expand my acronyms in future.

    Returning to my problem I did not understand how did you write [itex]x_2=\dfrac{x_1}{1+d}[/itex] and so on.
     
  5. Nov 22, 2012 #4

    utkarshakash

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    I get [itex]x_2=\dfrac{x_1}{1+x_1d}[/itex] Looks like you missed the x1 in each of your denominator.
     
    Last edited: Nov 22, 2012
  6. Nov 22, 2012 #5

    Ray Vickson

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    Hint: if
    [tex]\frac{1}{x_i} - \frac{1}{x_{i+1}} = r [/tex] for i = 1,2, ..., n-1 we have
    [tex]x_2 - x_1 = r x_1 x_2, \: x_3 - x_2 = r \, x_2 x_3, \ldots .[/tex] What does that give you?

    RGV
     
    Last edited: Nov 22, 2012
  7. Nov 22, 2012 #6

    haruspex

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    You know that d = 1/x2-1/x1, right? So what does that turn into for x1x2?
     
  8. Nov 23, 2012 #7

    utkarshakash

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    Thanks. It solved my problem.
     
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