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Math Help - Determine the progression?

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

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

    Answer:
    Spoiler:
    x_1 = \frac{a\sqrt{n + 1} - \sqrt{2(n - 1)(nb - a^2)}}{n\sqrt{n + 1}}, common difference = \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 x_1,x_2,x_3,\mbox{...},x_n are in Arithmetic Progression. It is given that x_1 + x_2 + \mbox{...} + x_n = a and x_1^2 + x_2^2 + \mbox{...} + x_n^2 = b^2. Determine the progression.

    Answer:
    Spoiler:
    x_1 = \frac{a\sqrt{n + 1} - \sqrt{2(n - 1)(nb - a^2)}}{n\sqrt{n + 1}}, common difference = \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 d. So the terms in the sequence are

     <br />
x_1, x_2 = x_1 + d, \cdots x_i = x_1 + (i-1)d,<br />

    Substitute in both expressions and expand. Formula's like

     <br />
\sum_{i=1}^n i = \frac{n(n+1)}{2},\;\;\; <br />
\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} <br />

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

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

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