A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
[c] \[\because \frac{b+c-2a}{b},\frac{c+a-2b}{a},\frac{a+b-2c~}{c}\]are in A.P. \[\Rightarrow \frac{2(c+a-2b)}{b}=\frac{b+c-2a}{a}+\frac{{{a}^{2}}+ab-2ca}{c}\] \[=\frac{bc+{{c}^{2}}-2ca+{{a}^{2}}+ab-2ca}{ca}\] \[\Rightarrow 2{{c}^{2}}a+2c{{a}^{2}}-4abc={{b}^{2}}+b{{c}^{2}}-4abc+{{a}^{2}}b+a{{b}^{2}}\]\[\Rightarrow 2ca\left( c+a \right)=be\left( b+c \right)+ab\left( a+b \right)\] Dividing a, b, c on both sides, \[\Rightarrow \frac{2\left( c+a \right)}{b}=\frac{b+c}{a}+\frac{a+b}{c}\] \[\Rightarrow \frac{2\left( c+a \right)~}{b}~+2=\frac{~b+c}{a}+1~\frac{a+b}{c}+1\] \[\Rightarrow \frac{2\left( a+b+c \right)}{b}=\frac{a+b+c}{a}=\frac{a+b+c}{c}\] \[\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}\] be in A.P. Hence a, b and c be in H.P.
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