If the roots of the equation $4x^3-12x^2+11x+k=0$  are in A.P then $k=$

Given roots are in A.P

$a-d,a,a+d$

$a-d+a+a+d=\frac{12}{4}$

$3a=\frac{12}{4}$

$a=1$

Then one of the root is $a=1$

Substitute $a$  in $f\left(x\right)$

$4-12+11+k=0$

$k=-3$