If a - bxa - bx - b - cxb - cx - c - dxc - dx x - 0- then a-b-c-d are in

The correct option is B G.P.From 1st two, by cross multiplication, we get #160; 2x (b^2 ac) = 0 ⇒ #160; b^2 ac = 0 as x ≠ 0 Sim ...

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If a + bx...

Question

If $\frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x} (x \neq 0),$ then a.b.c,d are in

A.P.

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G.P.

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H.P.

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none of these

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Similar questions

\frac{a - b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x} (x \neq 0)

a, b, c, d

If $\frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x} (x \neq 0),$ then show that a, b, c and d are in G.P.

If $\frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x} (x \neq 0)$ , then show that a, b, c and d are in G.P.

\frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x} (x \neq 0)

a, b, c

d

G . P .

If $\frac{a + b x}{a - b x}$ = $\frac{b + c x}{b - c x}$ = $\frac{c + d x}{c - d x}$ (x ≠ 0), then a, b, c, d are in:

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