If x, y, z are in A.P., then the value of the determinant a+2    a+3    a+2xa+3    a+4    a+2ya+4    a+5    a+2z is

Since x, y, z are in A.P., therefore, x + z - 2y = 0. Now,a+2    a+3    a+2xa+3    a+4    a+2ya+4    a+5    a+2z=002(x+z−2y)a+3a+4a+2ya+4a+5a+2z              Applying R1→R1+R3−2R2=000a+3a+4a+2ya+4a+5a+2z                  [∵x+z−2y=0]=0