If (y - x),2(y - a),and (y - z) are in H.P.,t | vedclass

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(B) Given that $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$This implies that their reciprocals $\frac{1}{y - x}$,$\frac{1}{2(y - a)}$,and $\frac{1

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$G.P.$

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(B) Given that $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$This implies that their reciprocals $\frac{1}{y - x}$,$\frac{1}{2(y - a)}$,and $\frac{1}{y - z}$ are in $A.P.$Therefore,$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$.Adding $\frac{1}{2(y - a)}$ to both sides,we get $\frac{1}{y - x} + \frac{1}{y - z} = \frac{2}{2(y - a)} = \frac{1}{y - a}$.$\frac{(y - z) + (y - x)}{(y - x)(y - z)} = \frac{1}{y - a}$.$(2y - x - z)(y - a) = (y - x)(y - z) = y^2 - yz - xy + xz$.$2y^2 - 2ay - xy + ax - zy + az = y^2 - yz - xy + xz$.$y^2 - 2ay + ax + az - xz = 0$.This simplifies to $(y - a)^2 = (x - a)(z - a)$.Thus,$(x - a)$,$(y - a)$,and $(z - a)$ are in $G.P.$

If $(y - x)$,$2(y - a)$,and $(y - z)$ are in $H.P.$,then $(x - a)$,$(y - a)$,and $(z - a)$ are in

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