If (y - x), 2(y - a), and (y - z) are in Harm | vedclass

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

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$G.P.$

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(B) Given that $(y - x), 2(y - a), (y - z)$ are in $H.P.$This implies that their reciprocals $\frac{1}{y - x}, \frac{1}{2(y - a)}, \frac{1}{y - z}$ are in $A.P.$Therefore,the common difference is equal:$\frac{1}{2(y - a)} - \frac{1}{y - x} = \frac{1}{y - z} - \frac{1}{2(y - a)}$$\frac{2}{2(y - a)} = \frac{1}{y - x} + \frac{1}{y - z}$$\frac{1}{y - a} = \frac{(y - z) + (y - x)}{(y - x)(y - z)}$$(y - x)(y - z) = (y - a)(2y - x - z)$$y^2 - yz - xy + xz = 2y^2 - xy - yz - 2ay + ax + az$$y^2 - xz = 2ay - ax - az$$y^2 - 2ay + a^2 = xz - ax - az + a^2$$(y - a)^2 = (x - a)(z - a)$This is the condition for $(x - a), (y - a), (z - a)$ to be in $G.P.$

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

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