If a, b, c are in G.P. (a 1, b 1, c 1), | vedclass

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(C) Given that $a, b, c$ are in $G$.$P$.,we have $b^2 = ac$. Taking the logarithm on both sides with base $x$, we get $2 \log_x b = \log_x a + \log_x

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$H$.$P$.

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(C) Given that $a, b, c$ are in $G$.$P$.,we have $b^2 = ac$. Taking the logarithm on both sides with base $x$, we get $2 \log_x b = \log_x a + \log_x c$. This implies that $\log_x a, \log_x b, \log_x c$ are in $A$.$P$. Using the change of base formula, $\log_a x = \frac{1}{\log_x a}$, $\log_b x = \frac{1}{\log_x b}$, and $\log_c x = \frac{1}{\log_x c}$. Since the reciprocals of $\log_x a, \log_x b, \log_x c$ are in $A$.$P$.,the terms $\log_a x, \log_b x, \log_c x$ must be in $H$.$P$.

If $a, b, c$ are in $G$.$P$. $(a > 1, b > 1, c > 1)$, then for any real number $x$ $(x > 0, x \neq 1)$, the terms $\log_{a} x, \log_{b} x, \log_{c} x$ are in:

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