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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.
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We have $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$
Applying componendo and dividendo, we get
$\frac{(a+b x)+(a-b x)}{(a+b x)-(a-b x)}=\frac{(b+c x)+(b-c x)}{(b+c x)-(b-c x)}$
$=\frac{(c+d x)+(c-d x)}{(c+d x)-(c-d x)}$
or $\quad \frac{2 a}{2 b x}=\frac{2 b}{2 c x}=\frac{2 c}{2 d x}$
or $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
Hence, $a, b, c, d$ are in G.P.
Applying componendo and dividendo, we get
$\frac{(a+b x)+(a-b x)}{(a+b x)-(a-b x)}=\frac{(b+c x)+(b-c x)}{(b+c x)-(b-c x)}$
$=\frac{(c+d x)+(c-d x)}{(c+d x)-(c-d x)}$
or $\quad \frac{2 a}{2 b x}=\frac{2 b}{2 c x}=\frac{2 c}{2 d x}$
or $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
Hence, $a, b, c, d$ are in G.P.
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