if x,y ,z are in A.P. then the value of the determinant
`|{:(a+2,,a+3,,a+2x),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2x):}|` is

Options

A. 1
B. 0
C. 2a
D. a

Answer

B

Explanation

It is given that x, y, z are in A.P.
`:. x + z - 2y = 0`
Now,
`|(a +2,a + 3,a + 2x),(a +3,a +4,a + 2y),(a + 4,a +5,a + 2z)|`
`= |(0,0,2(x + z - 2y)),(a + 3,a +4,a + 2y),(a +4,a +5,a + 2z)| " " ["Applying " R_(1) rarr R_(1) + R_(3) - 2R_(2)]`
`= |(0,0,0),(a +3,a +4,a + 2y),(a +4,a +5,a + 2z)| = 0 " " [ :' x + z - 2y = 0]`