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.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{a\mathit{+}bx}{a\mathit{-}bx}\mathit{=}\frac{b\mathit{+}cx}{b\mathit{-}cx}\mathit{=}\frac{c\mathit{+}dx}{c\mathit{-}dx} \\ \mathrm{Now},\frac{a\mathit{+}bx}{a\mathit{-}bx}\mathit{=}\frac{b\mathit{+}cx}{b\mathit{-}cx} \\ \mathrm{Applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo} \\ \Rightarrow \frac{\left(a\mathit{+}bx\right)\mathit{+}\left(a\mathit{-}bx\right)}{\left(a\mathit{+}bx\right)\mathit{-}\left(a\mathit{-}bx\right)}\mathit{=}\frac{\left(b\mathit{+}cx\right)\mathit{+}\left(b\mathit{-}cx\right)}{\left(b\mathit{+}cx\right)\mathit{-}\left(b\mathit{-}cx\right)} \\ \Rightarrow \frac{\mathit{2}a}{\mathit{2}bx}\mathit{}\mathit{=}\mathit{}\frac{\mathit{2}b}{\mathit{2}cx} \\ \Rightarrow \frac{a}{b}\mathit{=}\frac{b}{c} \\ \mathrm{Similiarly},\frac{\left(b\mathit{+}cx\right)\mathit{+}\left(b\mathit{-}cx\right)}{\left(b\mathit{+}cx\right)\mathit{-}\left(b\mathit{-}cx\right)}\mathit{=}\frac{\left(c\mathit{+}dx\right)\mathit{+}\left(c\mathit{-}dx\right)}{\left(c\mathit{+}dx\right)\mathit{-}\left(c\mathit{-}dx\right)} \\ \Rightarrow \frac{b}{c}\mathit{}\mathit{=}\mathit{}\frac{c}{d} \\ \mathrm{Therefore},a\mathit{,}\mathit{}b\mathit{,}\mathit{}c\mathrm{and}d\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \end{gathered}$$