If 2(y - a) is the H.M. between y - x and y - | vedclass

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(B) Given that $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$,therefore $y - x, 2(y - a), y - z$ are in $H.P.$This implies that $\frac{1}{y - x}

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$G.P.$

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(B) Given that $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$,therefore $y - x, 2(y - a), y - z$ are in $H.P.$This implies that $\frac{1}{y - x}, \frac{1}{2(y - a)}, \frac{1}{y - z}$ are in $A.P.$Thus,$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$.$\frac{2(y - a) - (y - x)}{2(y - a)(y - x)} = \frac{2(y - a) - (y - z)}{2(y - a)(y - z)}$$\frac{y - 2a + x}{y - x} = \frac{y - 2a + z}{y - z}$Rearranging terms: $\frac{y - 2a + x}{y - x} = \frac{y - 2a + z}{y - z}$By simplifying,we get $\frac{x - a + y - a}{-(x - a) + (y - a)} = \frac{y - a + z - a}{-(y - a) + (z - a)}$.Applying componendo and dividendo,we obtain $\frac{x - a}{y - a} = \frac{y - a}{z - a}$.Therefore,$x - a, y - a, z - a$ are in $G.P.$

If $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$,then $x - a, y - a, z - a$ are in

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