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$a, a + d, a + 2d, ……, a + 2nd$ શ્રેણીનું વિચરણ શોધો.
$\frac{{n(n\, + \,\,1)}}{2}\,{d^2}$
$\frac{{n(n\, + \,\,1)}}{3}\,{d^2}$
$\frac{{n(n\, + \,\,1)}}{6}{d^2}$
$\frac{{n(n\, + \,\,1)}}{{12}}\,{d^2}$
Solution
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}\, \times \,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{6}\,\, = \,\,\frac{{n(n\,\, + \,\,1)\,(2n\,\, + \,\,1)}}{3}\,\,{d^2}$
$\therefore \,\,\,Variance\,\, = \,\,\frac{{\Sigma {{({x_i}\, – \,\,\bar x)}^2}}}{{2n\,\, + \,\,1}}\,\, = \,\,\frac{{n(n\,\, + \,\,1)}}{3}\,{d^2}$
