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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
none of these
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Solution

The correct option is A A.P.
x1,y1,zGY
y2=x×z
taking log both sides
lny2=ln(x×z)
2lny=lnx+lnz
2(1+lny)=(1+lnx)+(1+lnz)
1+lnx,1+lny,1+lnz are in AP
Hence 11+lnx,11+lny,11+lnz are in H.P.

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