If frac{a + bx}{a - bx} = frac{b + cx}{b - cx | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = k$.Applying Componendo and Dividendo to each term:$\frac{(a + bx) +

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(B) Given $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx} = k$.Applying Componendo and Dividendo to each term:$\frac{(a + bx) + (a - bx)}{(a + bx) - (a - bx)} = \frac{(b + cx) + (b - cx)}{(b + cx) - (b - cx)} = \frac{(c + dx) + (c - dx)}{(c + dx) - (c - dx)}$.This simplifies to:$\frac{2a}{2bx} = \frac{2b}{2cx} = \frac{2c}{2dx}$.Dividing by $\frac{2}{x}$ (since $x 
eq 0$):$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$.This implies that $\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$.Therefore,$a, b, c, d$ are in $G.P.$

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c, d$ are in

Similar Questions

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

Which term of the $GP$ $2, 8, 32, \ldots$ is $131072$ (in $^{\text{th}}$)?

If $a, b, c, d$ are in a geometric progression,then:

If the roots of the equation $8x^3 - 14x^2 + 7x - 1 = 0$ are in $G.P.$,then the roots are

$1 - \frac{2}{3} + \frac{2 \cdot 4}{3 \cdot 6} - \frac{2 \cdot 4 \cdot 6}{3 \cdot 6 \cdot 9} + \ldots \infty =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module