If 12(y−a) is the arithmetic mean of 1y−x and 1y−z where a,x,y,z∈R, then x−a,y−a and z−a are in
A
A.P.
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B
G.P.
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C
A.G.P.
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D
H.P.
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Solution
The correct option is B G.P. Given : 12(y−a) is the arithmetic mean of 1y−x and 1y−z So, 1y−a=1y−x+1y−z ⇒(y−x)(y−z)=[2y−(x+z)](y−a)⇒y2−(x+z)y+xz=2y2−(x+z)y−2ay+a(x+z)⇒y2−2ay=xz−a(x+z)⇒y2−2ay+a2=xz−ax−az+a2⇒(y−a)2=(x−a)(z−a)