Question

If the sides a, b , c of a triangle ABC are in A.P. then $\frac{b}{c}$ belongs to –

• A. $\left( 0,\frac{2}{3} \right)$
• B. (1, 2)
• C. $\left( \frac{2}{3},2 \right)$
• D. $\left( \frac{2}{3},\frac{7}{3} \right)$

Answer 1

Answer: (C)

$\left( \frac{2}{3},2 \right)$

Answer 2

a + b > c ̃ 2b – c + b > c

̃ $\frac{b}{c} > \frac{2}{3}$

Also, b + c > a ̃ b + c > 2b – c

̃ $\frac{b}{c} < 2$

Again c + a > b ̃ 2b > b

\ $\frac{b}{c} \in \left( \frac{2}{3},2 \right)$