In $\triangle \mathrm{ABC}$, if $\frac{1}{\mathrm{r}_1}, \frac{1}{\mathrm{r}_2}$ and,…

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 $\mathrm{r}_2: \mathrm{r}=$
  1. $3: 2$
  2. $2: 1$
  3. $1: 3$
  4. $3: 1$

Solution

$\because \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ 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, $\frac{r_2}{r_1}=\frac{\frac{\Delta}{s-b}}{\frac{\Delta}{s}}=\frac{s}{s-b}=\frac{2 s}{2 s-2 b}=\frac{a+b+c}{a+b+c-2 b}$ $=\frac{2 b+b}{a+c-b}=\frac{3 b}{2 b-b}=\frac{3 b}{b} \Rightarrow \frac{r_2}{r}=\frac{3}{1}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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