Question

The geometric mean of the observations $x_1,x_2,x_3,.......x_n$ is $G_1$. The geometric mean of the observations $y_1,y_2,y_3,.....y_n$ is $G_2$. The geometric mean of observations $\frac{x_1}{y_1},\frac{x_2}{y_2},\frac{x_3}{y_3},......\frac{x_n}{y_n}$ is

• A. $G_1{G}_2$
• B. $\ln \left(G_1{G}_2\right)$
• C. $\frac{G_1}{G_2}$
• D. $\ln \left(\begin{array}{c}\frac{G_1}{G_2}\end{array}\right)$

Answer 1

The correct option is C $\frac{G_1}{G_2}$

Solution:

Given;

Geometric mean of $x_1,x_2,x_3,............x_n$ is $G_1.$

$⟹G_1=(x_1{x}_2{x}_3......x_n{)}^{\frac{1}{n}}$

Geometric mean of $y_1,y_2,y_3,............y_n$ is $G_2.$

$⟹G_2=(y_1{y}_2{y}_3......y_n{)}^{\frac{1}{n}}$

$\therefore GM$ of $\frac{x_1}{y_1},\frac{x_2}{y_2},\frac{x_3}{y_3},..............\frac{x_n}{y_n}$

$={(\frac{x_1}{y_1}\frac{x_2}{y_2}\frac{x_3}{y_3}..............\frac{x_n}{y_n})}^{\frac{1}{n}}=\frac{(x_1{x}_2{x}_3............x_n{)}^{\frac{1}{n}}}{(y_1{y}_2{y}_3..............y_n{)}^{\frac{1}{n}}}=\frac{G_1}{G_2}$

Hence, C is the correct option.