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

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

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$G.P.$

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(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = k$.Applying Componendo and Dividendo to each term:$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$.This simplifies to:$\frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx}$.Dividing by $\frac{2}{x}$ (since $x 
eq 0$):$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.This implies that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.Therefore,$a, b, c, d$ are in $G.P.$

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c, d$ are in

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