Question

If the determinant
$\left|\begin{array}{ccc}xp+y& x& y\\ yp+z& y& z\\ 0& xp+y& yp+z\end{array}\right|=0$ and $x,y,z,p\in R^{+}$ then

• A. $\text{x,y,z are in A.P.}$
• B. $\text{x,y,z are in G.P.}$
• C. $\text{x,y,z are in H.P.}$
• D. $\text{xy,yz,zx are in A.P.}$

Answer 1

The correct option is B $\text{x,y,z are in G.P.}$

$\text{Given}\left|\begin{array}{ccc}xp+y& x& y\\ yp+z& y& z\\ 0& xp+y& yp+z\end{array}\right|=0$;

$\text{Operating}{C}_1\to C_1-p{C}_2-C_3$
$\left|\begin{array}{ccc}0& x& y\\ 0& y& z\\ -(p^{2}x+2py+z)& xp+y& yp+z\end{array}\right|=0$

$\Rightarrow -(p^{2}x+2py+z)(xz-y^2)=0$
$\{\because p,x,y,z\in R^{+}\Rightarrow p^{2}x+2py+z\ne 0\}$

$\Rightarrow xz-y^2=0$

$\Rightarrow xz=y^2$

Hence, $x,y,z$ are in G.P.