if `a_(1),a_(2),a+_(3)……,a_(12)` are in A.P and
`Delta_(1)= |{:(a_(1)a_(5),,a_(1),,a_(2)),(a_(2)a_(6),,a_(2),,a_(3)),(a_(3)a_(7),,a_(3),,a_(4)):}| Delta_(3)= |{:(a_(2)b_(10),,a_(2),,a_(3)),(a_(3)a_(11),,a_(3),,a_(4)),(a_(3)a_(12),,a_(4),,a_(5)):}|`
then `Delta_(2):Delta_(2)= "_____"`

Answer

1

Explanation

`Delta_(1)= |{:(a_(1)^(2)+4a_(1)d,,a_(1),,d),(a_(2)^(2)+4a_(2)d,,a_(2),,d),(a_(3)^(2)+4a_(3)d,,a_(3),,d):}|, [C_(3)to C_(3)-C_(2)]`
where d is the common difference of A.P.
`= |{:(a_(1)^(2),,a_(1),,d),(a_(2)^(2),,a_(2),,1),(a_(3)^(2),,a_(3),,1):}|+4d |{:(a_(1),,a_(1),,d),(a_(2),,a_(2),,d),(a_(3),,a_(3),,d):}|`
`= d(a_(1) -a_(2))(a_(2)-a_(3))(a_(3)-a_(1))=-2d^(4)`
`"Similarly," Delta_(2) =-2d^(4)`