If $$\bar{X}$$ is the mean of $$x_1, x_2, x_3, \dots, x_n$$, then for $$a \neq 0$$,  the mean of $$(a x_1, a x_2, a x_3, \dots, a x_n$$, $$\frac{x_1}{a}$$, $$\frac{x_2}{a}$$, $$\frac{x_3}{a}$$,....$$\frac{x_n}{a}$$) is:

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Correct option is B

Given:
Let the arithmetic mean of x₁, x₂, x₃, ..., xₙ be X̄ 
Solution:
X̄  = $$\frac {x₁ + x₂ + x₃ + ... + xₙ} n$$ 
$$x₁ + x₂ + x₃ + ... + xₙ = n \bar{X}$$​​
 $$\begin{aligned}\text {Required Mean} &= \frac{(ax_1 + ax_2 + \ldots + ax_n) + \left( \frac{x_1}{a} + \frac{x_2}{a} + \ldots + \frac{x_n}{a} \right)}{n + n} \\[6pt]&= \frac{a(x_1 + x_2 + \ldots + x_n) + \frac{1}{a}(x_1 + x_2 + \ldots + x_n)}{2n} \\[6pt]&= \frac{(a + \frac{1}{a})(x_1 + x_2 + \ldots + x_n)}{2n} \\[6pt]&= \frac{(a + \frac{1}{a}) n \bar{X}}{2n} \\&= \frac{(a + \frac{1}{a}) \bar{X}}{2} \\\end{aligned}$$​​