If frac{{a + bx}}{{a - bx}} = frac{{b + cx}}{ | vedclass

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(b) $\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}$

Applying componendo and dividendo, we get

$\frac{{2a}}{{

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$G.P.$

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(b) $\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}$

Applying componendo and dividendo, we get

$\frac{{2a}}{{2bx}} = \frac{{2b}}{{2cx}} = \frac{{2c}}{{2dx}}$

$ \Rightarrow $${b^2} = ac$ and ${c^2} = bd$

$ \Rightarrow $$a,\;b,\;c$ and $b,\;c,\;d$ are in $G.P.$

Therefore, $a,\;b,\;c,\;d$ are in $G.P.$

If $\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}(x \ne 0)$, then $a,\;b,\;c,\;d$ are in

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