Question

The determinant is $\left|\begin{array}{ccc}xp+y& x& y\\ yp+z& y& z\\ 0& xp+y& yp+z\end{array}\right|=0$ if

• A. $x,y,z$ are in $AP$
• B. $x,y,z$ are in $GP$
• C. $x,y,z$ are in $HP$
• D. $xy,yz,zx$ are in $AP$

Answer 1

Answer: (B)

We have, $\left|\begin{array}{ccc}xp+y& x& y\\ yp+z& y& z\\ 0& xp+y& yp+z\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}xp& x& y\\ yp& y& z\\ -yp-z& xp+y& yp+z\end{array}\right|=0$
$[$Applying $C_1\to C_1-C_3]$
$\Rightarrow \left|\begin{array}{ccc}0& x& y\\ 0& y& z\\ -z-x{p}^2& xp+y& yp+z\end{array}\right|=0$
$[$ Applying $C_1\to C_1-p{C}_2]$
$\Rightarrow 0-0+(-z-x{p}^2)[xz-y^2]=0$
$[$ Expanding along $C_1]$
$\Rightarrow -(z-x{p}^2)[xz-y^2]=0$
$\Rightarrow z-x{p}^2=0$ or $xz-y^2=0$
$\therefore y^2=xz$
Hence, $x,y,z$ are in $GP$.