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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

$x = y = z$

Vedclass · Answer

(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

Similar Questions

The derivative of ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ with respect to ${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ is

The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to $1$ are

The number of positive integral solutions of $\tan ^{-1} x+\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$ are

Let $\cos ^{-1}(x) + \cos ^{-1} (2x) + \cos ^{-1}(3x) = \pi.$ If $x$ satisfies the cubic equation $ax^3 + bx^2 + cx - 1 = 0,$ then the value of $(a + b + c)$ is -

$\sin \left[ \cos^{-1} \left( \frac{3}{5} \right) + \tan^{-1} 2 \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module