Question

If x > 1, y > 1, z > 1 are in G.P., then $\frac{1}{1+lnx},\frac{1}{1+lny},\frac{1}{1+lnz}$ are in

• A. A.P
• B. G.P
• C. H.P
• D. A.G.P

Answer 1

The correct option is C H.P

As $x,y,z$ are in G.P we can rewrite them as:
$x=x,y=xk,z=x{k}^2$
Now considering the sequence $\frac{1}{1+\ln x},\frac{1}{1+\ln y},\frac{1}{1+\ln z}$
By substituting the values $x=x,y=xk,z=x{k}^2$ in the above equation the series is clearly in A.P.. with a common difference of $\ln k$.
So the given sequence is in H.P