Question

Question 13 If (a+bx) / (a-bx) = (b+cx) / (b-cx) = (c+dx) / (c-dx) x ≠ 0, then show that a, b, c and d are in G.P.

Class X1 - Maths -Sequences and Series Page 199

Answer 1

x is not zero, otherwise the statement is trivial (1=1=1) 
From 
(a+bx)/(a-bx) = (b+cx)/(b-cx) 
when we multiply both sides by both denominators and expand the products we get 

ab + (b^2)x - acx - bcx^2 = ab + acx - (b^2)x - bcx^2 
Hence 
2(b^2)x - 2acx = 0 
2x(b^2 - ac) = 0 

Therefore, since x is not zero, 
b^2 - ac = 0 
which is the condition for a, b, c to be in geometric progression, as b/a = c/b is equivalent to b^2 = ac. 

In the same way, b, c, d are in geometric progression. 
i.e. a, b, c, d are in geometric progression.

Answer 2

Given,
(a + bx)/(a - bx) = (b + cx)/(b - cx)
Applying componendo and dividendo,
if a/b = c/d , (a + b)/(a - b) = (c + d)/(c - d)

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

2a/2bx = 2b/2cx = 2c/2dx

a/b = b/c = c/d { common ratio is always constant }

Therefore,
a , b , c and d are GP