Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

x, y and z are in AP.

then, 2y = x+z ----(1)

${ \tan }^{ - 1} x \: , \: { \tan }^{ - 1} y \: and \: { \tan }^{ - 1} z \: are \: also \: in \: AP.$

then,

$2 { \tan}^{ - 1} y = { \tan }^{ - 1} x + { \tan }^{ - 1} z$

$→ { \tan }^{ - 1} ( \frac{2y}{1 - {y}^{2} } ) = { \tan}^{ - 1} ( \frac{x + z}{1 - xz} )$

$→ \frac{2y}{1 - {y}^{2} } = \frac{x + z}{1 - xz}$

from eqn(1)

$→ \frac{x + z}{1 - {y}^{2} } = \frac{x + z}{1 - xz}$

$→ \frac{1}{1 - {y}^{2} } = \frac{1}{1 - xz}$

$→1 - xz = 1 - {y}^{2}$

$→ - xz = - {y}^{2}$

$→ {y}^{2} = xz$

again, put value of y from eqn (1)

$→ {( \frac{x + z}{2} })^{2} = xz$

$→ \frac{ ({x + z})^{2} }{4} = xz$

$→ {x}^{2} + {z}^{2} + 2xz = 4xz$

$→ {x}^{2} + {z}^{2} + 2xz - 4xz = 0$

$→ {x}^{2} + {y}^{2} - 2xz = 0$

$→ {(x - z)}^{2} = 0$

$→x - z = 0$

$→x = z$

now, from eqn(1)..

$2y = x + x$

$→2y = 2x$

$→y = x$

→ x = y = z

Answer 2

$\rm x , y \: and \: z \: are \: in \: A.P \\ \\ \therefore \rm 2y = x + z...(1)\\ \\ \rm {tan}^{ - 1} x ,{tan}^{ - 1} y \: and \: {tan}^{ - 1} z \: are \: also \: in \: A.P \\ \\ \therefore \rm 2{tan}^{ - 1} y = {tan}^{ - 1} x +{tan}^{ - 1} z...(2)$

$\rm \longrightarrow 2{tan}^{ - 1} y = {tan}^{ - 1} x +{tan}^{ - 1} z \\ \\ \rm \longrightarrow {tan}^{ - 1} \bigg(\frac{2y}{1 - {y}^{2} } \bigg)={tan}^{ - 1} \bigg(\frac{x + z}{1 - x z } \bigg)\\ \\ \rm \longrightarrow \bigg(\frac{2y}{1 - {y}^{2} } \bigg)= \bigg(\frac{x + z}{1 - x z } \bigg)$

Now, 2y = x+z

$\rm \longrightarrow \bigg(\dfrac{x + z}{1 - {y}^{2} } \bigg)= \bigg(\dfrac{x + z}{1 - x z } \bigg) \\ \\ \rm \longrightarrow \dfrac{1 - x z}{1 - {y}^{2} } = \dfrac{x + z}{ x + z } \\ \\ \rm \longrightarrow \dfrac{1 - x z}{1 - {y}^{2} } = 1\\ \\ \rm \longrightarrow {1 - x z} = 1 - {y}^{2}\\ \\ \rm \longrightarrow { - x z} = - {y}^{2}\\ \\ \rm \longrightarrow { x z} = {y}^{2}...(3)$

➡️2y = x+z

➡️(2y)² = (x+z)²

➡️4y² = x²+z²+ 2xz

y²= xz

➡️4xz= x²+z²+ 2xz

➡️ 4xz - x²- z² - 2xz =0

➡️ 2xz - x²- z² =0

➡️ -2xz + x² +z² =0

➡️ (x -z)² =0

➡️ x -z =0

➡️ x = z

➝ 2y = x+z

put x = z

➝ 2y = z+z

➝ 2y = 2z

➝ y = z

Therefore, x= y = z

Also, from equation (3) we can see that x,y,z are in G.P and in the question it is given that x,y,z are in A.P.

If x, y,z are in both G.P and A.P then x=y=z

Answer :

Option (a) x=y=z is correct