Question

If $x,y,z$ are in arithmetic progression and $ta{n}^{-1}x,ta{n}^{-1}y$ and $ta{n}^{-1}z$ are also in arithmetic progression, then

• A. $x=y=z$
• B. $x=y=-z$
• C. $x=1,y=2,z=3$
• D. $x=2,y=4,z=6$

Answer 1

Answer: (A)

Since, $x,y,z$ are in AP
$\therefore y=\frac{x+z}{2}$ ....(i)
And $ta{n}^{-1}x,ta{n}^{-1}y$ and $ta{n}^{-1}z$ are also in AP.
$\therefore 2ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}z$
$\Rightarrow {\left(tan\right)}^{-1}\left(\frac{2y}{1-y^2}\right)={\left(tan\right)}^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow \frac{2y}{1-y^2}=\frac{2y}{1-xz}$ [from equation. (i)]
$\Rightarrow y^2=xz$
$\Rightarrow x,y,z$ are in GP.
$\therefore x=y=z$