a_1 , a_2 ,...,a_n are mutually distinct and is a am sequence . if a_( 1) +a_( 2) +...+a_n =A and a_1 ^( 2) + a_2 ^2 +..+ a


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Question Number 193199 by mnjuly1970 last updated on 07/Jun/23

$a_{1}, a_{2}, . . ., a_{n} a r e m u t u a l l y d i s t i n c t$ $a n d i s a a m s e q u e n c e .$ $i f a_{1} + a_{2} + . . . + a_{n} = A$ $a n d$ $a_{1}^{2} + a_{2}^{2} + . . + a_{n}^{2} = B$ $f i n d t h e a m s e q u e n c e .$
	Answered by MM42 last updated on 07/Jun/23

	$l e t x_{1} = a & x_{i} - x_{i - 1} = d$ $\Rightarrow A = n a + \frac{n (n - 1) d}{2} \Rightarrow \frac{A^{2}}{n} = {n a}^{2} + n (n - 1) a d + \frac{n {(n - 1)}^{2} d^{2}}{4} (i)$ $x_{i}^{2} = {(a + (i - 1) d)}^{2} = a^{2} + 2 a d (i - 1) + {(i - 1)}^{2} d^{2}$ $\Rightarrow \underset{i = 1}{\sum^{n}} x_{i}^{2} = {n a}^{2} + a d n (n - 1) + \frac{n (n - 1) (2 n - 1)}{6} d^{2} (i i)$ $(i i) - (i) \Rightarrow \frac{n B - A^{2}}{n} = \frac{n (n^{2} - 1)}{12} d^{2}$ $\Rightarrow d = \pm \frac{2 \sqrt{3 (n B - A^{2})}}{n \sqrt{n^{2} - 1}} ✓$
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