If a^x = b^y = c^z and a, b, c are in G.P.,th | vedclass

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(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac \dots (i)$.Let $a^x = b^y = c^z = k$.Then $a = k^{1/x}, b = k^{1/y}, c = k^{1/z}$.Substitutin

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$H.P.$

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(C) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac \dots (i)$.Let $a^x = b^y = c^z = k$.Then $a = k^{1/x}, b = k^{1/y}, c = k^{1/z}$.Substituting these values into equation $(i)$:$k^{2/y} = k^{1/x} \cdot k^{1/z} = k^{(1/x + 1/z)}$.Equating the exponents,we get $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.This implies that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$Therefore,$x, y, z$ are in $H.P.$

If $a^x = b^y = c^z$ and $a, b, c$ are in $G.P.$,then $x, y, z$ are in

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