A) A.P.
B) H.P.
C) G.P.
D) None of these
Correct Answer: B
Solution :
\[x,\ y,\ z\] are in G.P. Hence \[{{y}^{2}}=xz\] \[\therefore \]\[2\log y=\log x+\log z\] \[\Rightarrow \]\[2(\log y+1)=(1+\log x)+(1+\log z)\] \[\Rightarrow \]\[1+\log x,\ 1+\log y,\ 1+\log z\] are in A.P. \[\Rightarrow \]\[\frac{1}{1+\log x},\ \frac{1}{1+\log y},\ \frac{1}{1+\log z}\] are is H.P.
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