If then show that a, b, c and d are in G.P.

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.