Given {a^x} = {b^y} = {c^z} = {d^u} and a, b, | vedclass

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(C) Let ${a^x} = {b^y} = {c^z} = {d^u} = k$ (say).Then $a = {k^{1/x}}, b = {k^{1/y}}, c = {k^{1/z}}, d = {k^{1/u}}$.Since $a, b, c, d$ are in $G.P.$,t

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$H.P.$

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(C) Let ${a^x} = {b^y} = {c^z} = {d^u} = k$ (say).Then $a = {k^{1/x}}, b = {k^{1/y}}, c = {k^{1/z}}, d = {k^{1/u}}$.Since $a, b, c, d$ are in $G.P.$,their logarithms are in $A.P.$ or we can use the property of $G.P.$: ${b^2} = ac$ and ${c^2} = bd$.Substituting the values,we get ${k^{2/y}} = {k^{1/x}} \cdot {k^{1/z}} = {k^{(1/x + 1/z)}}$.This implies $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$,which means $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$Therefore,$x, y, z$ are in $H.P.$Similarly,for $b, c, d$ in $G.P.$,we get $\frac{2}{z} = \frac{1}{y} + \frac{1}{u}$,implying $y, z, u$ are in $H.P.$Thus,$x, y, z, u$ are in $H.P.$

Given ${a^x} = {b^y} = {c^z} = {d^u}$ and $a, b, c, d$ are in $G.P.$,then $x, y, z, u$ are in

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