Question

Which of the following statements is/are correct if $a,b\&c$ are three distinct natural numbers in H.P.?

• A. $ac\ge b$
• B. $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.
• C. $\frac{(c-1)}{2c}+\frac{(a-1)}{2a}=1-\frac{1}{b}$
• D. $a+c>b$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.
C $\frac{(c-1)}{2c}+\frac{(a-1)}{2a}=1-\frac{1}{b}$
D $a+c>b$

$G.M.>H.M.$
$\sqrt[3]{3abc}>bac>b^2$

$a,b,c$ are in $H.P.$
$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $A.P.$
$\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}=\frac{b+c-a}{a}+2,\frac{c+a-b}{b}+2,\frac{a+b-c}{c}+2$
$\Rightarrow \frac{b+c+a}{a},\frac{c+a+b}{b},\frac{a+b+c}{c}$ are in $A.P.$
$\Rightarrow \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in $A.P.$


$\frac{(c-1)}{2c}+\frac{(a-1)}{2a}=\frac{2ac-2a+2ac-2c}{4ac}=\frac{4ac-2a-2c}{4c}=1-\frac{1}{2}\frac{a+c}{ac}=1-\frac{1}{b}$


$A.M.>H.M.$
$\frac{a+b+c}{3}>ba+c>2b\Rightarrow a+c>b$