When frac{1}{a} + frac{1}{c} + frac{1}{a - b} | vedclass

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

Question

(A) Given the equation: $\frac{1}{a} + \frac{1}{c} + \frac{1}{a - b} + \frac{1}{c - b} = 0$Rearranging the terms,we get: $\frac{1}{a} + \frac{1}{c - b

Vedclass · Accepted Answer

in $H.P.$

Vedclass · Answer

(A) Given the equation: $\frac{1}{a} + \frac{1}{c} + \frac{1}{a - b} + \frac{1}{c - b} = 0$Rearranging the terms,we get: $\frac{1}{a} + \frac{1}{c - b} = -\frac{1}{c} - \frac{1}{a - b} = \frac{1}{b - a} - \frac{1}{c}$$\frac{c - b + a}{a(c - b)} = \frac{c - (b - a)}{(b - a)c}$$\frac{a + c - b}{ac - ab} = \frac{c - b + a}{bc - ac}$Assuming $a + c - b 
e 0$,we can cancel the term $(a + c - b)$ from both sides:$\frac{1}{ac - ab} = \frac{1}{bc - ac}$$bc - ac = ac - ab$$2ac = ab + bc$Dividing both sides by $abc$: $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$This is the condition for $a, b, c$ to be in $H.P.$

When $\frac{1}{a} + \frac{1}{c} + \frac{1}{a - b} + \frac{1}{c - b} = 0$ and $b \ne a \ne c$,then $a, b, c$ are

Similar Questions

If $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$,then

If $\tan B = \frac{2 \sin A \sin C}{\sin (A+C)}$,then $\tan A, \tan B$ and $\tan C$ are in

In a $H.P.$,the $p^{th}$ term is $q$ and the $q^{th}$ term is $p$. Then the $(pq)^{th}$ term is

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

The $4^{th}$ term of a $H.P.$ is $\frac{3}{5}$ and $8^{th}$ term is $\frac{1}{3},$ then its $6^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module