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

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(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

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in $H.P.$

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(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

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