If x,y,z are in A.P. and tan−1x, tan−1y, tan−1z are also in A.P., then
Solutions · 1
Official solution
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]
\[ \Rightarrow 1 - {y^2} = 1 - xz\]
\[ \Rightarrow \] y2=xz
As we get y2=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.