If `x , y , z` are in A.P. , then the value of the determinant are in A.P. , then the value of the determinant `|a+2a+3a+2x a+3a+4a+2y a+4a+5a+2z|` is a. 1 b. `0` c. `2a` d. `a`

Options

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

Answer

B

Explanation

since x,y,z are in A.P. therefore `x+z-2y=0` Now
`|{:(a+2,,a+3,,a+2x),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2x):}|=|{:(0,,0,,2(x+z-2y)),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2z):}|`
Applying `R_(1) to R_(1) +R_(3) -2R_(2)`
`=|{:(0,,0,,0),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2z):}|" "[ :' x +z -2y =0]`
`=0`