Question

If sides of triangle ABC are a,b and c such that $2b=a+c$, then exhaustive range of $\frac{b}{c}$ is

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

Answer 1

Answer: (C)

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

$\begin{array}{c}2b=a+c\\ \because a+c>b\\ anda-c<b\\ \therefore 2b-c+b>c\\ 3b>2c\\ \frac{b}{c}>\frac{2}{3}\\ so,a-c<b\\ 2b-c-c<b\\ b<2c\\ \frac{b}{c}<2\\ \frac{b}{c}\in \left(\frac{2}{3},2\right)\\ OptionCiscorrectanswer.\end{array}$