If \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\

1.	If \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\) (x \(\neq\) 0) then show that a, b, c and d are in G.Pa+bx /a - bx = b + cx/b - cx = c + dx/ c - dx (x ≠ 0)
Answer» Given : \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\) To Prove : a, b, c, and d are in G.P Proof : Applying component and dividend to the given expression, we get, \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\) \(\frac{a} {bx} = \frac{b}{cx} = \frac{c}{dx}\) \(\frac{a}{b} = \frac{b}{a} = \frac{c}{d}\) Clearly, a, b, c and d are in G.P. Hence proved.

If \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\) (x \(\neq\) 0) then show that a, b, c and d are in G.Pa+bx /a - bx = b + cx/b - cx = c + dx/ c - dx (x ≠ 0)

Answer»

Given : \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\)

To Prove : a, b, c, and d are in G.P

Proof :

Applying component and dividend to the given expression, we get,

\(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\)

\(\frac{a} {bx} = \frac{b}{cx} = \frac{c}{dx}\)

\(\frac{a}{b} = \frac{b}{a} = \frac{c}{d}\)

Clearly, a, b, c and d are in G.P.

Hence proved.