If {x^a} = {x^{b/2}}{z^{b/2}} = {z^c},then a, | vedclass

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(C) Let ${x^a} = {x^{b/2}}{z^{b/2}} = {z^c} = \lambda$. From this,we have $x = \lambda^{1/a}$,$z = \lambda^{1/c}$,and $xz = \lambda^{2/b}$. Substituti

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$H.P.$

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(C) Let ${x^a} = {x^{b/2}}{z^{b/2}} = {z^c} = \lambda$. From this,we have $x = \lambda^{1/a}$,$z = \lambda^{1/c}$,and $xz = \lambda^{2/b}$. Substituting the values of $x$ and $z$ into the expression for $xz$: $\lambda^{1/a} \cdot \lambda^{1/c} = \lambda^{2/b}$. Using the laws of exponents,we get $\lambda^{(1/a) + (1/c)} = \lambda^{2/b}$. Equating the exponents,we have $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$. This is the condition for $a, b, c$ to be in $H.P.$

If ${x^a} = {x^{b/2}}{z^{b/2}} = {z^c}$,then $a, b, c$ are in

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