# Determine the progression?

• May 21st 2009, 10:00 AM
fardeen_gen
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.

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?
• May 21st 2009, 11:11 AM
Jester
Quote:

Originally Posted by fardeen_gen
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.

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 $\displaystyle x_1$ .