A) 0
B) 1
C) 2
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Given that \[a,\ b,\ c\] are in G.P. So, \[{{b}^{2}}=ac\] ?..(i) \[x=\frac{a+b}{2}\] ?..(ii) \[y=\frac{b+c}{2}\] ?..(iii) Now \[\frac{a}{x}+\frac{c}{y}=\frac{2a}{a+b}+\frac{2c}{b+c}=\frac{2(ab+bc+2ca)}{ab+ac+{{b}^{2}}+bc}\] \[=\frac{2(ab+bc+2ca)}{(ab+ac+ac+bc)}=2\],\[\left\{ \because \ {{b}^{2}}=ac \right\}\]. Trick: Let \[a=1,\ b=2,\ c=4,\] then obviously \[x=\frac{3}{2}\] and\[y=3\], then \[\frac{1}{3/2}+\frac{4}{3}=2\].
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