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Q.
If $a, b, c$ are in H.P., then :
Sequences and Series
Solution:
(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 $=\frac{2 a c}{a+c}$
$\text { now } \left.\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2 a c}{a+c}-a}+\frac{1}{\frac{2 a c}{a+c}-c}=\frac{a+c}{c-a}\left[\frac{1}{a}-\frac{1}{c}\right]=\frac{a+c}{a c}=\frac{2}{b}=\text { L.H.S. }\right]$
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{a c}{a+c}\right)\left(c-\frac{a c}{a+c}\right) $
$=\frac{a c(a c)}{(a+c)^2}=\left(\frac{a c}{a+c}\right)^2=\left(\frac{b}{2}\right)^2 ; \text { hence } a-\frac{b}{2}, \frac{b}{2}, c-\frac{b}{2} \text { are in G.P. }$
(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. } $
$\Rightarrow \frac{1}{ a }, \frac{1}{ b }, \frac{1}{ c } \text { are in A.P. }$ $\Rightarrow a , b , c \text { in H.P. which is true ] }$