If x, y, z are in A.P., then the value of the determinant
$\begin{vmatrix} a+ 2& a+3& a+ 2x\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix},$ is
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : 0
It is given that $x, y, z $ are in A.P.
$∴x+z-2y = 0 $
Now,
$\begin{vmatrix} a+ 2& a+3& a+ 2x\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}$
$\begin{vmatrix} 0& 0 & 2(x+z-2y)\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}$ [Applying $R_1→R_1+R_3-2R_2$]
$=\begin{vmatrix} 0& 0 & 0\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}=0$ $[∵x+z-2y =0]$