Question

Statement-I : If $a,b,c$ are three distinct positive number in H.P., then $\left(\frac{a+b}{2a-b}\right)+\left(\frac{c+b}{2c-b}\right)>4$ Because
Statement-II : Sum of any number and it's reciprocal is always greater than or equal to 2.

• A. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I.
• B. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.
• C. Statement-I is true, Statement-II is false
• D. Statement-I is false, Statement-II is true.

Answer 1

Answer: (C)

st-I$:\frac{\frac{1}{b}+\frac{1}{a}}{\frac{2}{b}-\frac{1}{a}}+\frac{\frac{1}{b}+\frac{1}{c}}{\frac{2}{b}-\frac{1}{c}}=\frac{\frac{1}{b}+\frac{1}{a}}{\frac{1}{c}}+\frac{\frac{1}{b}+\frac{1}{c}}{\frac{1}{a}}\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\right)$
$=\frac{c}{b}+\frac{c}{a}+\frac{a}{b}+\frac{a}{c}=\frac{a+c}{b}+\left(\frac{c}{a}+\frac{a}{c}\right)$
$=\frac{(a+c{)}^2}{2ac}+\left(\frac{c}{a}+\frac{d}{c}\right)=\frac{1}{2}\left(\frac{d}{c}+\frac{c}{a}+2\right)+\left(\frac{c}{a}+\frac{d}{c}\right)>4$
$[\because x+\frac{1}{x}>2$ when $x>0,x\ne \frac{1}{x}]$
St.-II is False
$\therefore$ Numbers should be positive