If `x, y, z` are in A.P. and `tan^(-1)x , tan^(-1)y` and `t a n^(-1)z` are also in A.P., then
(1) `2x""=""3y""=""6z` (2) `6x""=""3y""=""2z` (3) `6x""=""4y""=""3z` (4) `x""=""y""=""z`

Options

A. `2x=3y=6z`
B. `6x=3y=2z`
C. `6x=4y=3z`
D. `x=y=z`

Answer

D

Explanation

`:.` x,y,z are in AP.
Let `x=y-d,z=y+d " " "…….(i)"`
Also, given `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP.
`:.2tan^(-1)y=tan^(-1)x+tan^(-1)z`
`implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))`
`implies (2y)/(1-y^(2))=(x+z)/(1-xz) implies (2y)/(1-y^(2))=(2y)/(1-(y^(2)-d^(2)))`
`implies y^(2)=y^(2)-d^(2)" " [" fromEq.(i) "]`
`:. d=0`
From Eq. (i), x=y and z=y
`:. x=y=z`
Aliter
`:. x,y,z" are in AP. "" " "…….(i)"`
`:. 2y=x+z" " "........(ii)"`
Also, `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in AP.
`:.2tan^(-1)y=tan^(-1)x+tan^(-1)z`
`implies tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))`
`implies (2y)/(1-y^(2))=(x+z)/(1-xz)=(2y)/(1-xz)" "[" from Eq.(ii) "]`
`implies y^(2)=zx`
`:. x,y,z " are in GP. "" "".........(iii)"`
From Eqs. (i) and (ii) x,y,z are in AP and also in GP, then x=y=z.