If the exradii of a triangle are in $H P$ then the corresponding sides are in

A P

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G P

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H P

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none of these

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Solution

The correct option is A $A P$
given that, $r_{1}, r_{2}, r_{3}$ are in H.P
Therefore, $\frac{2}{r_{2}} = \frac{1}{r_{1}} + \frac{1}{r_{3}}$
$\Rightarrow \frac{2 (s - b)}{△} = \frac{(s - a)}{△} + \frac{(s - c)}{△}$
$\Rightarrow 2 b = a + c$
Therefore, $a, b, c$ are in A.P
Ans: A

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