If the sides of a triangle are in A.P as well as in G.P., then the value of $\frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}}$ is

1

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0

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2 r

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N o n e

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Solution

The correct option is B $0$
$2 b = a + c, b^{2} = a c$
$∴ {(\frac{a + c}{2})}^{2} = a c \Rightarrow \frac{1}{4} {(a - c)}^{2} = 0$
$∴ a = c$
$∴ 2 b = 2 a$ as
$a = b = c$
$r_{1} = \frac{S}{s - a}, r_{2} = \frac{S}{s - b}, r_{3} = \frac{S}{s - c}$
$∴ r_{1} = r_{2} = r_{3} ∴ \frac{r_{1}}{r_{2}} - \frac{r_{2}}{r_{3}} = 1 - 1 = 0$

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