A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
\[2b=\frac{a+b}{1-ab}+\frac{b+c}{1-bc}\]\[\Rightarrow 2a{{b}^{3}}c+2abc=a+c+a{{b}^{2}}+{{b}^{2}}c\]\[\Rightarrow 2abc\,\,({{b}^{2}}+1)=(a+c)(1+{{b}^{2}})\]\[\Rightarrow 2b=\frac{a+c}{ac}\Rightarrow 2b=\frac{1}{a}+\frac{1}{c}\]\[\Rightarrow \]\[{{a}^{-1}},\,\]\[b,\]\[{{c}^{-1}}\] are in A.P. |
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