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In $\triangle \mathrm{ABC}$, if $\frac{1}{\mathrm{r}_1}, \frac{1}{\mathrm{r}_2}$ and, $\frac{1}{\mathrm{r}_3}$ are in arithmetic progression, then $\mathrm{r}_2: \mathrm{r}=$
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$3: 1$
$\because \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in arithmetic progression Then, $\frac{2}{r_2}=\frac{1}{r_1}+\frac{1}{r_3}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2(s-b)}{\Delta-}=\frac{s-a}{\Delta}+\frac{s-c}{\Delta} \\
& \Rightarrow 2(s-b)=2 s-(a+c) \\
& \Rightarrow 2 s-2 b=2 s-(a+c) \Rightarrow 2 b=a+c
\end{aligned}
$$
Now, $\frac{r_2}{r_1}=\frac{\frac{\Delta}{s-b}}{\frac{\Delta}{s}}=\frac{s}{s-b}=\frac{2 s}{2 s-2 b}=\frac{a+b+c}{a+b+c-2 b}$ $=\frac{2 b+b}{a+c-b}=\frac{3 b}{2 b-b}=\frac{3 b}{b} \Rightarrow \frac{r_2}{r}=\frac{3}{1}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2(s-b)}{\Delta-}=\frac{s-a}{\Delta}+\frac{s-c}{\Delta} \\
& \Rightarrow 2(s-b)=2 s-(a+c) \\
& \Rightarrow 2 s-2 b=2 s-(a+c) \Rightarrow 2 b=a+c
\end{aligned}
$$
Now, $\frac{r_2}{r_1}=\frac{\frac{\Delta}{s-b}}{\frac{\Delta}{s}}=\frac{s}{s-b}=\frac{2 s}{2 s-2 b}=\frac{a+b+c}{a+b+c-2 b}$ $=\frac{2 b+b}{a+c-b}=\frac{3 b}{2 b-b}=\frac{3 b}{b} \Rightarrow \frac{r_2}{r}=\frac{3}{1}$
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