If frac{1}{b - a} + frac{1}{b - c} = frac{1}{ | vedclass

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

Question

(C) Given: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$ Rearranging the terms: $\frac{1}{b - a} - \frac{1}{a} = \frac{1}{c} - \frac

Vedclass · Accepted Answer

$H.P.$

Vedclass · Answer

(C) Given: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$ Rearranging the terms: $\frac{1}{b - a} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b - c}$ $\frac{a - (b - a)}{a(b - a)} = \frac{(b - c) - c}{c(b - c)}$ $\frac{2a - b}{a(b - a)} = \frac{b - 2c}{c(b - c)}$ $(2a - b)(bc - c^2) = (b - 2c)(ab - a^2)$ $2abc - 2ac^2 - b^2c + bc^2 = ab^2 - a^2b - 2abc + 2a^2c$ $4abc - 2ac^2 - b^2c + bc^2 - ab^2 + a^2b - 2a^2c = 0$ This simplifies to $b(a+c) = 2ac$,which is the condition for $a, b, c$ to be in $H.P.$ Thus,$a, b, c$ are in $H.P.$

If $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$,then $a, b, c$ are in

Similar Questions

If $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are in $H.P.$,then

If $a, b, c$ are in $H.P.$,then $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in

If the sixth term of a $H.P.$ is $\frac{1}{61}$ and its tenth term is $\frac{1}{105},$ then the first term of that $H.P.$ is

If for the harmonic progression,$t_{7} = \frac{1}{10}$ and $t_{12} = \frac{1}{25}$,then $t_{20} =$

If $x, y, z$ are in harmonic progression,then $\log(x + z) + \log(x - 2y + z) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module