Question

Statement-I : If $a, b, c$ are three distinct positive number in H.P., then $\left(\frac{a+b}{2 a-b}\right)+\left(\frac{c+b}{2 c-b}\right)>4$ 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$: \frac{\frac{1}{b}+\frac{1}{a}}{\frac{2}{b}-\frac{1}{a}}+\frac{\frac{1}{b}+\frac{1}{c}}{\frac{2}{b}-\frac{1}{c}}=\frac{\frac{1}{b}+\frac{1}{a}}{\frac{1}{c}}+\frac{\frac{1}{b}+\frac{1}{c}}{\frac{1}{a}}\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\right)$
$=\frac{ c }{ b }+\frac{ c }{ a }+\frac{ a }{ b }+\frac{ a }{ c }=\frac{ a + c }{ b }+\left(\frac{ c }{ a }+\frac{ a }{ c }\right)$
$=\frac{( a + c )^2}{2 ac }+\left(\frac{ c }{ a }+\frac{ d }{ c }\right)=\frac{1}{2}\left(\frac{ d }{ c }+\frac{ c }{ a }+2\right)+\left(\frac{ c }{ a }+\frac{ d }{ c }\right)>4$
$\left[\because x+\frac{1}{x}>2\right.$ when $\left.x>0, x \neq \frac{1}{x}\right]$
St.-II is False
$ \therefore$ Numbers should be positive