a, a + d, a + 2d, ……, a + 2nd&n | vedclass

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Question

આપેલ શ્રેણી $\,a,\,\,a\,\, + \,\,d,\,\,a\,\, + \,\,2d,\,.......\,,\,\,a\,\, + \,\,2nd$

$\bar x\,\, = \,\,\,\frac{{\frac{{(2n\,\, + \,\,1)}}{2}

Vedclass · Accepted Answer

$\frac{{n(n\, + \,\,1)}}{3}\,{d^2}$

Vedclass · Answer

આપેલ શ્રેણી $\,a,\,\,a\,\, + \,\,d,\,\,a\,\, + \,\,2d,\,.......\,,\,\,a\,\, + \,\,2nd$

$\bar x\,\, = \,\,\,\frac{{\frac{{(2n\,\, + \,\,1)}}{2}\,[a\,\, + \,\,a\,\, + \,\,2nd]}}{{(2n\,\, + \,\,1)}}\,\,\, = \,\,a\,\, + \,\,nd$

$\Sigma {({x_i}\, - \,\,\bar x)^2}\, = \,\,{n^2}{d^2}\, + \,\,{(1\,\, - \,\,n)^2}{d^2}\, + \,\,.......\,\, + \,\,{d^2}\, + \,\,0\,\, + \,\,{d^2}\, + \,\,.......\,\, + \,\,{n^2}{d^2}$

$ = \,\,2[{n^2}{d^2}\, + \,\,{(n\,\, - \,\,1)^2}{d^2}\, + \,\,......\,\, + \,\,{d^2}]\,\,\,\,\,\,\,\,\,$

$ = \,\,2{d^2}[{n^2}\, + \,\,{(n\,\, - \,1)^2}\, + \,\,......\,\, + \,\,{1^2}]$

$ = \,\,2{d^2}\, 	imes \,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{6}\,\, = \,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{3}\,\,{d^2}$

$	herefore \,\,\,Variance\,\, = \,\,\frac{{\Sigma {{({x_i}\, - \,\,\bar x)}^2}}}{{2n\,\, + \,\,1}}\,\, = \,\,\frac{{n(n\,\, + \,\,1)}}{3}\,{d^2}$

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