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The A.M of $a, a+d, a+2d.....a+2nd$ is

$\begin {array} {1 1} (A)\;a+(n-1)d & \quad (B)\;a+(n+1)d \\ (C)\;a+(n+2)d & \quad (D)\;None \: of \: these \end {array}$

1 Answer

$ \overline x = \large\frac{a+(a+d)+(a+2d)+....+(a+2nd)}{2n+1}$
$ = \large\frac{1}{ \not{2n +1}}. \large\frac{\not{2n+1}}{2}$ $ ( a+a+2nd)$
$ = a+nd$
Ans : (D)
answered Jan 25, 2014 by thanvigandhi_1
 

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