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

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$2 x = 3 y = 6 z$

$x = y = z$

$6 x = 3 y = 2 z$

$6 x = 4 y = 3 z$

answer is A.

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Detailed Solution

We know that if x, y,z are in A.P then
$\begin{array}{l} \Rightarrow 2 y = x + z \\ G i v e n \tan^{- 1} x, \tan^{- 1} y, \tan^{- 1} z are in A.P \\ \Rightarrow 2 \tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} z \\ \Rightarrow \tan^{- 1} (\frac{2 y}{1 - y^{2}}) = \tan^{- 1} (\frac{x + z}{1 - xz}) \\ \Rightarrow \frac{2 y}{1 - y^{2}} = \frac{x + z}{1 - xz} \\ \Rightarrow \frac{x + z}{1 - y^{2}} = \frac{x + z}{1 - xz} (∵ 2 y = x + z) \\ 1 - y^{2} = 1 - xz \\ y^{2} = zx \Rightarrow x, y, z are in G.P \\ since x, y, z are in A.P and also in G.P then \\ x = y = z \end{array}$