Question

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

• A. H.P
• B. A.P
• C. G.P
• D. None

Answer 1

Answer: (A)

H.P

$x,y,z$ are in G.P, hence

$y^2=xz$

Taking log on both sides we get

$ln\left(y^2\right)=ln\left(xz\right)$

$\Rightarrow 2ln\left(y\right)=ln\left(x\right)+ln\left(z\right)$

$\Rightarrow 2+2ln\left(y\right)=ln\left(x\right)+ln\left(z\right)+2$

$\Rightarrow 1+1+2ln\left(y\right)=ln\left(x\right)+ln\left(z\right)+1+1$

$\Rightarrow 2(1+ln(y)=(ln\left(x\right)+1\left)\right)+(1+ln(z\left)\right)$

$\Rightarrow 2(1+ln(y\left)\right)=(1+ln(x\left)\right)+(1+ln(z\left)\right)$

Hence, $(1+ln(x\left)\right),(1+ln(y\left)\right),(1+ln(z\left)\right)$

are in AP.

So, $\frac{1}{1+ln\left(x\right)},\frac{1}{1+ln\left(y\right)},\frac{1}{1+ln\left(z\right)}$ are

in H.P

So, option (A) iis correct.