If $G_{1}, G_{2}$ are the geometric means of two series of observations and $G$ is the GM of the ratios of the corresponding observations then $G$ is equal to

\frac{G_{1}}{G_{2}}

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log G_{1} - log G_{2}

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\frac{log G_{1}}{log G_{2}}

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log (G_{1} . G_{2})

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Solution

The correct option is A $\frac{G_{1}}{G_{2}}$
Let $x_{1}, x_{2}, . . x_{n}$ be n obeservation corresponding to variate $x$ and $y_{1}, y_{2}, . . y_{n}$ be n observation corresponding to variate $y$
Then by defination
$G_{1} = {(x_{1} . x_{2} . . . x_{n})}^{\frac{1}{n}}$ and $G_{2} = {(y_{1} . y_{2} . . . y_{n})}^{\frac{1}{n}}$
Now the geometric mean of this is given by
$G = {(\frac{x_{1}}{y_{1}} . \frac{x_{2}}{y_{2}} . \frac{x_{3}}{y_{3}} . . . \frac{x_{n}}{y_{n}})}^{\frac{1}{n}} = {(\frac{x_{1} . x_{2} . . . x_{n}}{y_{1} . y_{2} . . . y_{n}})}^{\frac{1}{n}} = \frac{G_{1}}{G_{2}}$

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