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Question

If x>1,y>1,z>1 are three numbers in GP then
11+lnx, 11+lny, 11+lnz are in

A
AP
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B
HP
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C
GP
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D
none of these
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Solution

The correct option is B HP
Given y2=xz
2lny=lnx+lnz
lnx,lny,lnz are in AP
lnx+1,lny+1,lnz+1 are in AP
1lnx+1,1lny+1,1lnz+1 are in HP

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