Unusual setup

Adding the roots of the polynomial, We get $a+a+a+d-d=3a$

Now, From Vieta's Formulas, We know that Sum of roots of the Polynomial $= \dfrac{-b}{a}= \dfrac{-(-12)}{4}=3$

Therefore $3a=3, a=1.$

Now since,a=1 is a root of the polynomial, p(1) should be =0

$=3 \times 1-12 \times 1 +11 +k=0$

$3+k=0$

$k=(-3)$

Answer:- $\boxed{-3}$

Cheers!

Let a - d, a and a + d be the roots that forms an A.P. Since the sum of the roots is 3, 3a = 3 or a = 1 (one of the three roots). Just plug in 1 will give us k + 3 or k = -3.

Let the roots of the polynomial be $a$ , $b$ and $c$ and $a<b<c$ . Since they form an arithmetic progression with a common difference of $d$ , then $a = b-d$ and $c=b+d$ .

Now by Vieta's Formulas, we have:

$\begin{cases} a+b+c = \frac{12}{4} = 3 & \Rightarrow b-d + b + b + d = 3 \\ & \Rightarrow 3b = 3 \quad \Rightarrow b = 1 \\ ab+bc+ca = \frac{11}{4} & \Rightarrow (1-d)(1) +(1)(1+d) + (1+d)(1-d) \\ & \quad = 3-d^2 = \frac{11}{4} \quad \Rightarrow d^2 = \frac{1}{4} \quad \Rightarrow d = \pm \frac{1}{2} \\ & \Rightarrow a = \frac{1}{2} \quad b = 1 \quad c = \frac{3}{2} \\ abc = -\frac{k}{4} & \Rightarrow \frac{1}{2}\dot{} 1 \dot{} \frac{3}{2} = - \frac{k}{4} \quad \Rightarrow -\frac{k}{4} = \frac{3}{4} \quad \Rightarrow k = \boxed{-3} \end{cases}$