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Question

If a, b, c are in G.P. and a1/x=b1/y=c1/z, then x,y,z are in

A
A.P.
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B
G.P.
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C
H.P.
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D
Cannot be determined
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Solution

The correct option is A A.P.
We have,
a1/x=b1/y=c1/z=λ
a=λx,b=λy,c=λz
Now, a,b,c are in G.P.
b2=ac
(λy)2=λx×λz=λx+z
2y=x+z
x,y,z are in A.P.

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