Question

If $x, y$ and $z$ are in AP and $tan^{-1}x, tan^{-1} y$ and $tan^{-1} z $ are also in $AP$, then

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

Answer 1

Answer: (A)

Since, x , y and z are in an AP.
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, 2y=x+z$
Also, $tan^{-1} x , tan^{-1} y \, and \, tan^{-1} z$ are in an AP.
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2tan^{-1} y =tan^{-1} x+tan^{-1} (z) $
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, tan^{-1}\bigg(\frac{2y}{1-y^2}\bigg) = tan^{-1}\bigg(\frac{x+z}{1-xz}\bigg)$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{x+z}{1-y^2}=\frac{x+z}{1-xz}$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, y^2=xz$
Since x, y and z are in an AP a s well as in a GP.
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, x=y=z$