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If x>1, y>1, z>1 are in G.P., then 11+ln x, 11+ln y, 11+ln z are in 

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a
A.P.
b
G.P.
c
H.P.
d
none of these

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detailed solution

Correct option is C

As x, y, z are in G.P. ⇒y2=xz⇒ ln⁡ y2=ln⁡ (xz) ⇒ 2ln⁡ (y)=ln⁡ (x)+ln⁡ (z)⇒ 2(1+ln⁡ (y))=(1+ln⁡ (x))+(1+ln⁡ (z))⇒ 1+ln⁡ (x),1+ln⁡ y),1+ln⁡ (z) are in A.P. ⇒ 11+ln⁡ (x),11+ln⁡ (y),11+ln⁡ (z) are in H.P.


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