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

Options

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

Answer

A

Explanation

`2y = x + z`
As `tan^(-1)x, tan^(-1) y, tan^(-1) z` are in A.P., we have
`2 tan^(-1) y = tan^(-1) x + tan^(-1) z`
`rArr 2 tan^(-1) y = tan^(-1).(x+ z)/(1 -xz)`
`rArr tan^(-1).(2y)/(1 -y^(2)) = tan^(-1).(x + z)/(1 -xz)`
`rArr (2y)/(1 -y^(2)) = (x + z)/(1 -xz)`
`rArr (x + z)/(1 -y^(2)) = (x + z)/(1 -xz)`
`rArr (x +z) {(1)/(1 -y^(2)) -(1)/(1-xz)}`
`x + z = 0 " or " 1 -xz = 1 -y^(2)`
`rArr y^(2) = xz`
i.e., x, y, z are in G.P
Since x, y, z are in A.P. and G.P.
`:. x = y = z`