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If A.M of $x_1, x_2....x_n$ is $ \overline x$ then A.M of $ ax_1+b, ax_2+b, ax_3+b...ax_4+b$ is

$\begin {array} {1 1} (A)\;a\overline x & \quad (B)\;a\overline x+b \\ (C)\;a\overline x+nb & \quad (D)\;None\: of \: these \end {array}$

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1 Answer

A.M = $ \large\frac{(ax_1+b)+(ax_2+b)+....(ax_n+b)}{n}$
$ = \large\frac{a(x_1+x_2+....x_n)+nb}{n}$
$ = \large\frac{a(x_1+x_2+....+x_n)}{n}$$+b$
$ = a\overline x + b$
Ans : (B)
answered Jan 25, 2014 by thanvigandhi_1
 

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