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If the arithmetic mean of the numbers $x_1, x_2, x_3, \ldots . ., x_n$ is $\bar{x}$, then the arithmetic mean of numbers $a x_1+b, a x_2+b, a x_3+b, \ldots \ldots . ., a x_n+b$, where $a, b$ are two constants would be
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The correct answer is:
$a \bar{x}+b$
Required mean $=\frac{\left(a x_1+b\right)+\left(a x_2+b\right)+\ldots . .+\left(a x_n+b\right)}{n}$
$\begin{aligned} &=\frac{a\left(x_1+x_2+\ldots .+x_n\right)+n b}{n}=a \bar{x}+b, \\ & \quad\left(\because \frac{x_1+x_2+\ldots .+x_n}{n}=\bar{x}\right) .\end{aligned}$
$\begin{aligned} &=\frac{a\left(x_1+x_2+\ldots .+x_n\right)+n b}{n}=a \bar{x}+b, \\ & \quad\left(\because \frac{x_1+x_2+\ldots .+x_n}{n}=\bar{x}\right) .\end{aligned}$
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