If x, y, z are in Arithmetic Progression and | vedclass

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$ . . . $(i)$ Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A.P.$,so $2 	an^{-1} y =

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$x = y = z$

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(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$ . . . $(i)$ Also,$	an^{-1} x, 	an^{-1} y, 	an^{-1} z$ are in $A.P.$,so $2 	an^{-1} y = 	an^{-1} x + 	an^{-1} z$. Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} (\frac{a+b}{1-ab})$,we get $	an^{-1} (\frac{2y}{1-y^2}) = 	an^{-1} (\frac{x+z}{1-xz})$. This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$. Substituting $x+z = 2y$ from $(i)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$. Since $y Since $x, y, z$ are in both $A.P.$ and $G.P.$,it must be that $x = y = z$.

If $x, y, z$ are in Arithmetic Progression and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are also in Arithmetic Progression,where $x, z > 0$ and $xz < 1, y < 1$,then

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