If G_1 and G_2 are the geometric means of two | vedclass

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(D) Let the two series be $x_1, x_2, \dots, x_{n_1}$ and $y_1, y_2, \dots, y_{n_2}$ with sizes $n_1$ and $n_2$ respectively.$G_1 = (x_1 	imes x_2 	i

Vedclass · Accepted Answer

$\frac{n_1 \log G_1 + n_2 \log G_2}{n_1 + n_2}$

Vedclass · Answer

(D) Let the two series be $x_1, x_2, \dots, x_{n_1}$ and $y_1, y_2, \dots, y_{n_2}$ with sizes $n_1$ and $n_2$ respectively.$G_1 = (x_1 	imes x_2 	imes \dots 	imes x_{n_1})^{1/n_1} \implies G_1^{n_1} = x_1 	imes x_2 	imes \dots 	imes x_{n_1} \dots (1)$$G_2 = (y_1 	imes y_2 	imes \dots 	imes y_{n_2})^{1/n_2} \implies G_2^{n_2} = y_1 	imes y_2 	imes \dots 	imes y_{n_2} \dots (2)$The combined geometric mean $G$ is given by:$G = [(x_1 	imes x_2 	imes \dots 	imes x_{n_1}) 	imes (y_1 	imes y_2 	imes \dots 	imes y_{n_2})]^{\frac{1}{n_1 + n_2}}$Substituting $(1)$ and $(2)$ into the expression for $G$:$G = (G_1^{n_1} 	imes G_2^{n_2})^{\frac{1}{n_1 + n_2}}$Taking the logarithm on both sides:$\log G = \log [(G_1^{n_1} 	imes G_2^{n_2})^{\frac{1}{n_1 + n_2}}]$Using the property $\log(a^b) = b \log a$ and $\log(ab) = \log a + \log b$:$\log G = \frac{1}{n_1 + n_2} \log (G_1^{n_1} 	imes G_2^{n_2})$$\log G = \frac{n_1 \log G_1 + n_2 \log G_2}{n_1 + n_2}$

If $G_1$ and $G_2$ are the geometric means of two series of sizes $n_1$ and $n_2$ respectively,and $G$ is the geometric mean of their combined series,then what is $\log G$ equal to?

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