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If a, b and c are in G.P and x and y, respectively , be arithmetic means between a,b and b,c then prove that `a/x+c/y=2`and `1/x+1/y=2/b`

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a,b,c are in G.P. Hence,
`b^(2)=ac` (1)
x is A.M. of a and b. Hence,
2x=a+b (2)
y is A.M. of b and c. Hence,
2y=b+c (3)
`thereforea/x+c/y=axx2/(a+b)+cxx2/(b+c)` [Using (2) and (3)]
`=2[(ab+ac+ac+bc)/(ab+ac+b^(2)+bc)]`
=2 [Using (1)]
Again,
`1/x+1/y=2/(a+b)+2/(b+c)`
`=(2(a+c+2b))/(ab+ac+b^(2)+bc)`
`=(2(a+c+2b))/(ab+2b^(2)+bc) (becauseb^(2)=ac)`
`=(2(a+c+2b))/(b(a+c+2b))=2/b`
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