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Thread: Determine the progression?

  1. #1
    Super Member fardeen_gen's Avatar
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    Determine the progression?

    Suppose $\displaystyle x_1,x_2,x_3,\mbox{...},x_n$ are in Arithmetic Progression. It is given that $\displaystyle x_1 + x_2 + \mbox{...} + x_n = a$ and $\displaystyle x_1^2 + x_2^2 + \mbox{...} + x_n^2 = b^2$. Determine the progression.

    Answer:
    Spoiler:
    $\displaystyle x_1 = \frac{a\sqrt{n + 1} - \sqrt{2(n - 1)(nb - a^2)}}{n\sqrt{n + 1}}$, common difference = $\displaystyle \frac{2}{n}\sqrt{\frac{3(nb - a^2)}{(n^2 - 1)}}$


    I was unable to do it. Any ideas?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by fardeen_gen View Post
    Suppose $\displaystyle x_1,x_2,x_3,\mbox{...},x_n$ are in Arithmetic Progression. It is given that $\displaystyle x_1 + x_2 + \mbox{...} + x_n = a$ and $\displaystyle x_1^2 + x_2^2 + \mbox{...} + x_n^2 = b^2$. Determine the progression.

    Answer:
    Spoiler:
    $\displaystyle x_1 = \frac{a\sqrt{n + 1} - \sqrt{2(n - 1)(nb - a^2)}}{n\sqrt{n + 1}}$, common difference = $\displaystyle \frac{2}{n}\sqrt{\frac{3(nb - a^2)}{(n^2 - 1)}}$


    I was unable to do it. Any ideas?
    A tour de force. Let the common difference by $\displaystyle d$. So the terms in the sequence are

    $\displaystyle
    x_1, x_2 = x_1 + d, \cdots x_i = x_1 + (i-1)d,
    $

    Substitute in both expressions and expand. Formula's like

    $\displaystyle
    \sum_{i=1}^n i = \frac{n(n+1)}{2},\;\;\;
    \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
    $

    will be useful. You should end up with the following:

    $\displaystyle
    n x_1 + \frac{n(n-1)}{2} d = a,\;\;(1)
    $
    $\displaystyle
    nx_1^2 + x_1 n (n-1) d + \frac{n(n-1)(2n-1)}{6} d^2 = b
    $ (note $\displaystyle b$ and not $\displaystyle b^2$)

    Eliminate $\displaystyle x_1$ and solve for $\displaystyle d$ gives your answer. then use (1) to get $\displaystyle x_1$ .
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