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Question

If x > 1, y > 1, z > 1 are in G.P., then 11+lnx,11+lny,11+lnz are in

A
A.P
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B
G.P
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C
H.P
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D
A.G.P
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Solution

The correct option is C H.P
As x,y,z are in G.P we can rewrite them as:
x=x,y=xk,z=xk2
Now considering the sequence 11+lnx,11+lny,11+lnz
By substituting the values x=x,y=xk,z=xk2 in the above equation the series is clearly in A.P.. with a common difference of lnk.
So the given sequence is in H.P

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