Determine the progression?

**fardeen_gen** · May 21st 2009, 10:00 AM

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:

I was unable to do it. Any ideas?

**Danny** · May 21st 2009, 11:11 AM

Originally Posted by fardeen_gen

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:

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

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

Substitute in both expressions and expand. Formula's like

$ \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:

$ n x_1 + \frac{n(n-1)}{2} d = a,\;\;(1) $
$ nx_1^2 + x_1 n (n-1) d + \frac{n(n-1)(2n-1)}{6} d^2 = b $ (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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