Question

If $\Delta =\left|\begin{array}{cccc}3& 4& 5& x\\ 4& 5& 6& y\\ 5& 6& 7& z\\ x& y& z& 0\end{array}\right|=0$, then

• A. $x,y,z$ are in A.P
• B. $x,y,z$ are in G.P
• C. $x,y,z$ are in H.P
• D. none of these

Answer 1

The correct option is A $x,y,z$ are in A.P

Given $△=\left|\begin{array}{cccc}3& 4& 5& x\\ 4& 5& 6& y\\ 5& 6& 7& z\\ x& y& z& 0\end{array}\right|$

$R_1+R_3-2R_2\to R_1$

$\therefore △=\left|\begin{array}{cccc}0& 0& 0& x+z-2y\\ 4& 5& 6& y\\ 5& 6& 7& z\\ x& y& z& 0\end{array}\right|$

$\therefore △=-(x+z-2y)\left|\begin{array}{ccc}4& 5& 6\\ 5& 6& 7\\ x& y& z\end{array}\right|$

$C_1+C_3-2C_2\to C_1$

$\therefore △=-(x+z-2y)\left|\begin{array}{ccc}0& 5& 6\\ 0& 6& 7\\ x+z-2y& y& z\end{array}\right|$

$\therefore △=-(x+z-2y)(x+z-2y)(35-36)$

$\therefore △=-(x+z-2y)(x+z-2y)(-1)$

$\therefore △={(x+z-2y)}^2$

As given in the problem, $△=0$

$\therefore {(x+z-2y)}^2=0$

$\therefore x+z-2y=0$

$\therefore 2y=x+z$

$\therefore y=\frac{x+z}{2}$

Thus, $x,y,z$ are in A.P.