If a, b, c are in harmonic progression,then t | vedclass

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

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$(1, - 2)$

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(C) Given that $a, b, c$ are in harmonic progression $(H.P.)$,we have $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$,which implies $\frac{1}{a} + \frac{1}{c} - \frac{2}{b} = 0$ $... (i)$The given equation of the line is $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ $... (ii)$We can rewrite equation $(ii)$ as $\frac{1}{a}(x) + \frac{1}{c}(1) + \frac{1}{b}(y) = 0$.From equation $(i)$,we know $\frac{1}{c} = \frac{2}{b} - \frac{1}{a}$. Substituting this into the line equation:$\frac{x}{a} + \frac{y}{b} + (\frac{2}{b} - \frac{1}{a}) = 0$Rearranging the terms to group by $\frac{1}{a}$ and $\frac{1}{b}$:$\frac{1}{a}(x - 1) + \frac{1}{b}(y + 2) = 0$For this equation to hold for all $a, b, c$ in $H.P.$,the coefficients must be zero:$x - 1 = 0 \Rightarrow x = 1$$y + 2 = 0 \Rightarrow y = -2$Thus,the fixed point is $(1, -2)$.

If $a, b, c$ are in harmonic progression,then the straight line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ always passes through a fixed point. That point is:

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