Question

$a,b$ and $c$ are three distinct number in harmonic progression$;a,b,c>0;$ then which of the following option(s) is/are correct?

• A. $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P
• B. $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.
• C. $a^5+c^5\ge 2b^5$
• D. $\frac{a-b}{b-c}=\frac{a}{c}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P
B $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.
D $\frac{a-b}{b-c}=\frac{a}{c}$

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

(A) $\frac{a+b+c}{a},\frac{a+b+c}{b},\frac{a+b+c}{c}$ are in A.P
$\Rightarrow \frac{a+b+c}{a}-2,\frac{a+b+c}{b}-2,\frac{a+b+c}{c}-2$ are also in A.P
$\Rightarrow \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P

(B) Similarly, $\frac{a+b+c}{a}-1,\frac{a+b+c}{b}-1,\frac{a+b+c}{c}-1$ are in A.P
$\Rightarrow \frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.

(C) $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge \frac{2}{\sqrt{ac}}\Rightarrow \sqrt{ac}\ge b$
$\Rightarrow a^5+c^5\ge 2(ac{)}^{\frac{5}{2}}\ge 2b^5$

(D) $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$
$\Rightarrow 2ac=ab+bc$
$\Rightarrow \frac{a-b}{b-c}=\frac{a}{c}$