Question

If $x,y,z$ are in A.P., then the value of the determinant
$\left|\begin{array}{ccc}a+2& a+3& a+2x\\ a+3& a+4& a+2y\\ a+4& a+5& a+2z\end{array}\right|$ is

• A. 1
• B. 0
• C. 2a
• D. a

Answer 1

Answer: (B)

As $x,y,z$ are in A.P. Therefore, $x+z-2y=0$
Now, $\left|\begin{array}{ccc}a+2& a+3& a+2x\\ a+3& a+4& a+2y\\ a+4& a+5& a+2z\end{array}\right|$
$\left|\begin{array}{ccc}0& 0& 2\left(x+z-2y\right)\\ a+3& a+4& a+2y\\ a+4& a+5& a+2z\end{array}\right|$
[applying $R_1\to R_1+R_3-2R_2]$
$\left|\begin{array}{ccc}0& 0& 0\\ a+3& a+4& a+2y\\ a+4& a+5& a+2z\end{array}\right|$
$[\because x+z-2y=0]$
$=0$