Thank you for reporting, we will resolve it shortly
Q.
If the non-zero numbers $x, y, z$ are in A.P. and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in A.P., then
Sequences and Series
Solution:
Since $x, y, z$ are in $A.P . $
$ \therefore 2y = x+z \quad...\left(1\right)$
Since $tan^{-1}x, tan^{-1}y, tan^{-1}z$ and in $A.P.$
$\therefore 2tan^{-1}y = tan^{-1}x+tan^{-1}z $
$ tan^{-1} \frac{2y}{1-y^{2}} = tan^{-1}\frac{ x+z}{1-xz} $
$\Rightarrow \frac{2y}{1-y^{2}} =\frac{ x+z}{1-xz} $
$ \therefore x, y, z$ are in $H.P. $
Also $y^{2} = xz $
$ \Rightarrow x, y, z$ are in $G.P.$
$\therefore x= y=z $