Question

If $x,y$ and $z$ are in AP and $ta{n}^{-1}x,ta{n}^{-1}y$ and $ta{n}^{-1}z$ are also in $AP$, then

• A. $x=y=z$
• B. $2x=3y=6z$
• C. $6x=3y=2z$
• D. $6x=4y=3z$

Answer 1

Answer: (A)

Since, x , y and z are in an AP.
$\therefore 2y=x+z$
Also, $ta{n}^{-1}x,ta{n}^{-1}yandta{n}^{-1}z$ are in an AP.
$\therefore 2ta{n}^{-1}y=ta{n}^{-1}x+ta{n}^{-1}(z)$
$\Rightarrow ta{n}^{-1}\left(\frac{2y}{1-y^2}\right)=ta{n}^{-1}\left(\frac{x+z}{1-xz}\right)$
$\Rightarrow \frac{x+z}{1-y^2}=\frac{x+z}{1-xz}$
$\Rightarrow y^2=xz$
Since x, y and z are in an AP a s well as in a GP.
$\therefore x=y=z$