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If in a ΔABC, a, b, c are in A.P. then it is necessary that

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$\frac{2}{3} < \frac{b}{a} \geq 2$

$\frac{1}{3} < \frac{b}{c} < \frac{2}{3}$

$\frac{1}{3} < \frac{b}{a} < \frac{2}{3}$

$\frac{2}{3} < \frac{b}{c} < 2$

answer is A, C.

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Detailed Solution

a + c = 2b $\Rightarrow$ a + b > c $\Rightarrow$ b + c > a $\Rightarrow$ a + b > c
$\Rightarrow$ 3b > 2c $\Rightarrow$ b + c > a $\Rightarrow$ 2c > b
$\Rightarrow \frac{2}{3} < \frac{b}{c} < 2; Similarly for \frac{b}{a}$

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