QJEEM13Mafec6: JEE Main 2013 Mathematics - Class 12 JEEM13Mafec6 Solution: If are in A.P. and and are also in A.P., then :

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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.

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