JEE MAIN - Mathematics (2013 (Offline) - No. 8)
If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then :
$$x=y=z$$
$$2x=3y=6z$$
$$6x=3y=2z$$
$$6x=4y=3z$$
Explanation
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 $$x=y=z.$$
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 $$x=y=z.$$
Comments (0)
