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 $G.P.$
It is given that,
$\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$
$\Rightarrow(a+b x)(b-c x)=(b+c x)(a-b x)$
$\Rightarrow a b-a c x+b^{2} x-b c x^{2}=a b-b^{2} x+a c x-b c x^{2}$
$\Rightarrow 2 b^{2} x=2 a c x$
$\Rightarrow b^{2}=a c$
$\Rightarrow \frac{b}{a}=\frac{c}{b}$ .........$(1)$
Also, $\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$
$\Rightarrow(b+c x)(c-d x)=(b-c x)(c+d x)$
$\Rightarrow b c-b d x+c^{2} x-c d x^{2}=b c+b d x-c^{2} x-c d x^{2}$
$\Rightarrow 2 c^{2} x=2 b d x$
$\Rightarrow c^{2}=b d$
$\Rightarrow \frac{c}{d}=\frac{d}{c}$ .........$(2)$
From $(1)$ and $(2),$ we obtain
$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
Thus, $a, b, c$ and $d$ are in $G.P.$
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