In triangle ABC,if frac{1}{r_1}, frac{1}{r_2} | vedclass

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(D) Given that $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in arithmetic progression,we have $\frac{2}{r_2} = \frac{1}{r_1} + \frac{1}{r_3}$.Usi

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$3 : 1$

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(D) Given that $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in arithmetic progression,we have $\frac{2}{r_2} = \frac{1}{r_1} + \frac{1}{r_3}$.Using the formulas $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$ and $r = \frac{\Delta}{s}$,we get:$\frac{2(s-b)}{\Delta} = \frac{s-a}{\Delta} + \frac{s-c}{\Delta}$$2s - 2b = 2s - (a+c)$$2b = a+c$.This implies that $a, b, c$ are in arithmetic progression.Now,we need to find $r_2 : r = \frac{r_2}{r} = \frac{\Delta / (s-b)}{\Delta / s} = \frac{s}{s-b}$.Since $2b = a+c$,we have $s = \frac{a+b+c}{2} = \frac{3b}{2}$.Thus,$\frac{s}{s-b} = \frac{3b/2}{3b/2 - b} = \frac{3b/2}{b/2} = 3$.Therefore,$r_2 : r = 3 : 1$.

In $\triangle ABC$,if $\frac{1}{r_1}, \frac{1}{r_2}$ and $\frac{1}{r_3}$ are in arithmetic progression,then $r_2 : r =$

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