Given that, $x,y,z\,\,$ are in $AP$ So, $\,\,\,$ $2y = x + y$ Also given that, ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y\,\,$ and $\,\,\,{\tan ^{ - 1}}z\,\,$ are in $AP$ So, $2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$ $\Rightarrow {\tan ^{ - 1}}\left( {{{2y} \over {1 - {y^2}}}} \right) = {\tan ^{ - 1}}\left( {{{x + 7} \over {1 - xz}}} \right)$ $\Rightarrow {{2y} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$ $\Rightarrow {{x + z} \over {1 - {y^2}}} = {{x + z} \over {1 - xz}}$ [ as $\,\,\,\,$ $2y = x + z$] $\Rightarrow 1 - {y^2} = 1 - xz$ $\Rightarrow$ ${y^2} = xz$ As we get ${y^2} = xz,$ so, $x,y,z$ are in $GP.$ According to the question $x,y,z$ are $AP.$ $x,y,z$ both can be $AP$ as well as $GP$ when $x=y=z.$