Question

The determinant $\left|\begin{array}{c}xp+y\\ yp+z\\ 0\end{array}\begin{array}{c}x\\ y\\ xp+y\end{array}\begin{array}{c}y\\ z\\ yp+z\end{array}\right|=0$ for all $pϵR$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

The correct option is B x, y, z are in GP

$\left|\begin{array}{ccc}xp+y& x& y\\ yp+z& y& z\\ 0& xp+y& yp+z\end{array}\right|=0$
$C_1⟶C_1-{pC}_2-C_3$
$\left|\begin{array}{ccc}0& x& y\\ 0& y& z\\ {-xp}^2-yp-yp-z& xp+y& yp+z\end{array}\right|=0$
$(-{xp}^2-2py-z)(xz-y^2)=0$
$\therefore$ for all $pϵR$
${xp}^2+2yp+z=0$
is not possible
$\therefore y^2=xz$
$\therefore x,y,z$ are in G.P.