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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
H.P
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C
G.P
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D
none of above
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Solution

The correct option is A H.P
x,y,z are in GP

y2=xz

Taking log on both sides

2lny=lnx+lnz

2(1+lny)=(1+lnx)+(1+lnz)

i.e 1+lnx,1+lny,1+lnz are in AP

11+lnx,11+lny,11+lnz are in HP.

Hence, option B.

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