Question

If $x, \, y, \, z$ are in arithmetic progression and $tan^{- 1} x , tan^{- 1} ⁡ y$ and $tan^{- 1} z$ are also in arithmetic progression, then

• A. $x=y=z$
• B. $x=y=-z$
• C. $x=1, \, y=2, \, z=3$
• D. $x=2, \, y=4, \, z=6$

Answer 1

Answer: (A)

Since, $x, \, y, \, z$ are in AP
$∴ \, \, \, y=\frac{x + z}{2}$ ....(i)
And $tan^{- 1} x , tan^{- 1} ⁡ y$ and $tan^{- 1} z$ are also in AP.
$∴ \, \, \, 2tan^{- 1} y = tan^{- 1} ⁡ x + tan^{- 1} ⁡ z$
$⇒ \, \, \, \left(tan\right)^{- 1} \left(\frac{2 y}{1 - y^{2}}\right)=\left(tan\right)^{- 1} ⁡ \left(\frac{x + z}{1 - x z}\right)$
$⇒ \, \, \, \frac{2 y}{1 - y^{2}}=\frac{2 y}{1 - x z}$ [from equation. (i)]
$⇒ \, \, \, y^{2}=xz$
$⇒ \, \, \, x, \, y, \, z$ are in GP.
$∴ \, \, \, x=y=z$