Question

If $a,b,c$ are in H.P., then :

• A. $\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.
• B. $\frac{2}{b}=\frac{1}{b-a}+\frac{1}{b-c}$
• C. $a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}$ are in G.P.
• D. $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in H.P.

Answer 1

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

(A) $\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$ are in H.P.
$\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}\text{ are inA.P. }$
$\frac{a+b+c}{a},\frac{b+c+a}{b},\frac{a+b+c}{c}\text{ are in A.P. (adding }2\text{ in each term) }$
$\frac{1}{a},\frac{1}{b},\frac{l}{c}\text{ are in A.P. }\Rightarrow a,b,c\text{ are in }HP\Rightarrow A$
(B) a, b, c are in H.P. $\Rightarrow$ b $=\frac{2ac}{a+c}$
$\text{ now }\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2ac}{a+c}-a}+\frac{1}{\frac{2ac}{a+c}-c}=\frac{a+c}{c-a}\left[\frac{1}{a}-\frac{1}{c}\right]=\frac{a+c}{ac}=\frac{2}{b}=\text{ L.H.S. }]$
consider the number
(C)$a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2};\text{ now }\left(a-\frac{b}{2}\right)\left(c-\frac{b}{2}\right)=\left(a-\frac{ac}{a+c}\right)\left(c-\frac{ac}{a+c}\right)$
$=\frac{ac(ac)}{(a+c{)}^2}={\left(\frac{ac}{a+c}\right)}^2={\left(\frac{b}{2}\right)}^2;\text{ hence }a-\frac{b}{2},\frac{b}{2},c-\frac{b}{2}\text{ are in G.P. }$
(D) Let $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in H.P. is true
$\Rightarrow \frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}\text{ are in A.P. }\Rightarrow \frac{a+b+c}{a},\frac{b+c+a}{b},\frac{c+a+b}{c}\text{ in AP. }$
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P. }$ $\Rightarrow a,b,c\text{ in H.P. which is true ] }$