A) \[\frac{2}{b}\]
B) \[\frac{3}{b}\]
C) \[\frac{b}{3}\]
D) \[\frac{b}{2}\]
E) \[\frac{1}{b}\]
Correct Answer: A
Solution :
Given, \[{{b}^{2}}=ac,x=\frac{a+b}{2}\]and \[y=\frac{b+c}{2}\] \[\therefore \] \[\frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c}\] \[=\frac{2(2b+a+c)}{ab+{{b}^{2}}+bc+ac}\] \[=\frac{2(2b+a+c)}{ab+2{{b}^{2}}+bc}\] \[=\frac{2(2b+a+c)}{b(2b+a+c)}\] \[=\frac{2}{b}\]You need to login to perform this action.
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