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

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(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$.Applying componendo and dividendo to each term:$\frac{(a + bx) + (a

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(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$.Applying componendo and dividendo to each term:$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$$\Rightarrow \frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx}$$\Rightarrow \frac{a}{bx} = \frac{b}{cx} = \frac{c}{dx}$Since $x 
e 0$,we can cancel $x$ from the denominators:$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$This implies that the ratio of consecutive terms is constant.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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