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Question
If \[\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}\] (x ≠ 0), then show that a, b, c and d are in G.P.
Solution
\[\text { Given }: \]
\[\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}\]
\[\text { Now, } \frac{a + bx}{a - bx} = \frac{b + cx}{b - cx}\]
\[\text { Applying componendo and dividendo }\]
\[ \Rightarrow \frac{\left( a + bx \right) + \left( a - bx \right)}{\left( a + bx \right) - \left( a - bx \right)} = \frac{\left( b + cx \right) + \left( b - cx \right)}{\left( b + cx \right) - \left( b - cx \right)}\]
\[ \Rightarrow \frac{2a}{2bx} = \frac{2b}{2cx}\]
\[ \Rightarrow \frac{a}{b} = \frac{b}{c}\]
\[\text { Similiarly, } \frac{\left( b + cx \right) + \left( b - cx \right)}{\left( b + cx \right) - \left( b - cx \right)} = \frac{\left( c + dx \right) + \left( c - dx \right)}{\left( c + dx \right) - \left( c - dx \right)}\]
\[ \Rightarrow \frac{b}{c} = \frac{c}{d}\]
\[\text { Therefore, a, b, c and d are in G . P } .\]
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