If  $$\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}=\dfrac{c+dx}{c-dx}(x\neq0)$$, then show that a, b, c and d are in G.P.

Asked by Abhisek | 1 year ago |  59

##### Solution :-

Given,

$$\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}$$

$$= (a+bx)(b-cx)$$

$$=(b+cx)(a-bx)$$

$$ab-acx+b^2x-bc^2x$$

$$=ab-b^2x+-acx-bcx^2$$

$$2b^2x=2acx$$

$$b^2=ac$$

$$\dfrac{b}{a}=\dfrac{c}{b}$$

It is also given that,

$$\dfrac{b+cx}{b-cx}=\dfrac{c+dx}{c-dx}$$

$$=(b+cx)(c-dx)$$

$$=(b-cx)(c+dx)$$

$$bc-bdx+c^2x-cdx^2$$

$$=bc+bdx--c^2x-cdx^2$$

$$2c^2x=2bdx$$

$$c^2=bd$$

$$\dfrac{c}{d}=\dfrac{d}{c}$$

Equating both the results, we get

$$\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{b}$$

Therefore, it is proved that a,b,c and d are in G.P.

Answered by Pragya Singh | 1 year ago

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