(c)

We have, $p(x)=a x^2+b x+c$, where $a, b, c$ are in $AP$ and $a, b, c$ are positive real.

$\alpha, \beta$ are root of $p(x)=0$, where $\alpha$ and $\beta$ are integers.

$p(x) =a x^2+b x+c=0$

$\alpha+\beta =\frac{-b}{a}, \alpha \beta=\frac{c}{a}$

$\alpha, \beta$ are integer.

$\alpha+\beta=\frac{-b}{a}=-\lambda, \lambda \in I$

$b=a \lambda$

$a, b, c$ are in $AP$

$b=\frac{a+c}{2} \Rightarrow \frac{a+c}{2}=a \lambda$

$c=a(2 \lambda-1)$

$\therefore a x^2+a \lambda x+a(2 \lambda-1) =0$

$\Rightarrow x^2+\lambda x+(2 \lambda-1) =0 \quad[\because a \neq 0]$

$D=\lambda^2-4(2 \lambda-1)$ is a perfect square for integral roots.

$\lambda^2-8 \lambda+4 =k^2$

$(\lambda-4)^2-12 =k^2$

$\Rightarrow \quad(\lambda-4-k)(\lambda-4+k)=2 \times 6$

$\Rightarrow \lambda-4-k=2$ and $\lambda-4+k=6$

$\because \lambda=8$ and $k=2$

$\therefore \quad \alpha+\beta+\alpha \beta=\frac{-b}{a}+\frac{c}{a}$

$=\frac{-a \lambda+a(2 \lambda-1)}{a}$

$=\frac{a(\lambda-1)}{a}$

$=\lambda-1=8-1=7$