Question

If $\overline{x}$ is the mean of $x_1,x_2,...,x_n,$ then for $a\ne 0$, the mean of $a{x}_{1,}a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$ is
(a) $\left(a+\frac{1}{a}\right)\overline{x}$

(b) $\left(a+\frac{1}{a}\right)\frac{\overline{x}}{2}$

(c) $\left(a+\frac{1}{a}\right)\frac{\overline{x}}{n}$

(d) $\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\overline{x}}{2n}$

Answer 1

(b) $\left(a+\frac{1}{a}\right)\frac{\overline{x}}{2}$
$$\begin{gathered} \mathrm{Required}\mathrm{mean}=\frac{(a{x}_1+a{x}_2+...+a{x}_n)+\left({\displaystyle \frac{x_1}{a}}+{\displaystyle \frac{x_2}{a}}+...+{\displaystyle \frac{x_n}{a}}\right)}{2n} \\ =\frac{1}{2}\left\{\frac{a(x_1+x_2+...+x_n)}{n}+\frac{{\displaystyle \frac{1}{a}}(x_1+x_2+...+x_n)}{n}\right\} \\ =\frac{1}{2}\left\{a\overline{x}+\frac{1}{a}\overline{x}\right\}\left(\overline{x}=\frac{x_1+x_2+...+x_n}{n}\right) \\ =\left\{a+\frac{1}{a}\right\}\frac{\overline{x}}{2} \end{gathered}$$