If three non-zero real numbers a, b, c are in | vedclass

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(A) Given that $a, b, c$ are in harmonic progression,we have $\frac{1}{b} = \frac{1}{2} (\frac{1}{a} + \frac{1}{c})$,which implies $\frac{2}{b} = \fra

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$(1, -2)$

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(A) Given that $a, b, c$ are in harmonic progression,we have $\frac{1}{b} = \frac{1}{2} (\frac{1}{a} + \frac{1}{c})$,which implies $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$.Rearranging this,we get $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$.The given equation of the line is $\frac{x}{a} + \frac{y}{b} - \frac{2}{c} = 0$.Comparing this with the condition $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$,we can rewrite the line equation as $\frac{1}{a}(x) + \frac{1}{b}(y) + \frac{1}{c}(-2) = 0$.For this to be true for all $a, b, c$ satisfying the harmonic progression condition,the coefficients must satisfy the same linear relation.By inspection,if we set $x = 1$ and $y = -2$,the equation becomes $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$,which is exactly the condition for harmonic progression.Thus,the lines are concurrent at the point $(1, -2)$.

If three non-zero real numbers $a, b, c$ are in harmonic progression,then the straight lines $\frac{x}{a} + \frac{y}{b} - \frac{2}{c} = 0$ are concurrent at the point

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