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 rule to the first two parts:$\frac

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Geometric Progression

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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 rule to the first two parts:$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)}$$\frac{2a}{2bx} = \frac{2b}{2cx} \implies \frac{a}{b} = \frac{b}{c} \implies b^2 = ac$.Similarly,applying it to the last two parts:$\frac{2b}{2cx} = \frac{2c}{2dx} \implies \frac{b}{c} = \frac{c}{d} \implies c^2 = bd$.Since $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$,the terms $a, b, c, d$ are in Geometric Progression.

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

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