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) \[y=\frac{b+b}{2}\] ? (iii) Now \[\frac{x}{a}+\frac{c}{y}=\frac{2a}{a+b}+\frac{2c}{b+c}=\frac{2(ab+bc+2ca)}{ab+ac+{{b}^{2}}+bc}\] \[=\,\,\frac{2(ab+bc+2ca)}{(ab+ac+ac+bc)}=2\] \[\left[ \because \,\,{{b}^{2}}=ac \right]\]You need to login to perform this action.
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