# Thread: Determine the progression?

1. ## 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?

2. 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:
$x_1 = \frac{a\sqrt{n + 1} - \sqrt{2(n - 1)(nb - a^2)}}{n\sqrt{n + 1}}$, 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

$
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 $x_1$ .