Question

If x,y,z are the integers in A.P, lying between 1 and 9 and x51,y41 and z31 are three digits numbers, then the value of $\left|\begin{array}{ccc}5& 4& 3\\ x51& y41& z31\\ x& y& z\end{array}\right|$ is

• A. $x+y=z$
• B. $x+z=2y$
• C. $0$
• D. None of these

Answer 1

The correct option is B $x+z=2y$

$D=\left|\begin{array}{ccc}5& 4& 3\\ 100x+50+1& 100y+40+1& 100z+30+1\\ x& y& z\end{array}\right|$

$R_2\to R_2-100R_3-10R_1$

$=\left|\begin{array}{ccc}5& 4& 3\\ 100x+50+1-100x-50& 100y+40+1-100y-40& 100z+30+1-100z-30\\ x& y& z\end{array}\right|$

$=\left|\begin{array}{ccc}5& 4& 3\\ 1& 1& 1\\ x& y& z\end{array}\right|$

$\Rightarrow 5(z-y)-4(z-x)+3(y-x)=0$

$\Rightarrow 5z-5y-4z+4x+3y-3x=0$

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

$x+z=2y$ since $x,y,z$ are in A.P