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

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Question

(b) $y - x,\;2(y - a),\;(y - z)$ are in $H.P.$

$ \Rightarrow $ $\frac{1}{{y - x}},\;\frac{1}{{2(y - a)}},\;\frac{1}{{y - z}}$are in A.P.

$ \Righ

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(b) $y - x,\;2(y - a),\;(y - z)$ are in $H.P.$

$ \Rightarrow $ $\frac{1}{{y - x}},\;\frac{1}{{2(y - a)}},\;\frac{1}{{y - z}}$are in A.P.

$ \Rightarrow $ $\frac{1}{{2(y - a)}} - \frac{1}{{y - x}} = \frac{1}{{y - z}} - \frac{1}{{2(y - a)}}$

$ \Rightarrow $ $\frac{{2a - y - x}}{{y - x}} = \frac{{y + z - 2a}}{{y - z}}$

$ \Rightarrow $ $\frac{{(x - a)+ (y - a)}}{{(x - a) - (y -a)}} = \frac{{(y - a) + (z - a)}}{{(y - a) - (z - a)}}$

$ \Rightarrow $ $\frac{{x - a}}{{y - a}} = \frac{{y - a}}{{z - a}}$

(Applying componendo and dividendo)

$ \Rightarrow $ $x - a,\;y - a,\;z - a$ 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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