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

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(B) Given the equation $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = K$.Applying the property of Componendo and Dividendo,$

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$G.P.$

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(B) Given the equation $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = K$.Applying the property of Componendo and Dividendo,$\frac{(a+bx) + (a-bx)}{(a+bx) - (a-bx)} = \frac{2a}{2bx} = \frac{a}{bx}$.Similarly,for the other terms,we get $\frac{a}{bx} = \frac{b}{cx} = \frac{c}{dx}$.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,which is the definition of a Geometric Progression $(G.P.)$.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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