Need solutions

From Vieta's Formula $\alpha +\beta=-\dfrac{b}{a}$ and $\alpha \beta=\dfrac{c}{a}$

As $\alpha$ and $\beta$ are integers then $a|b$ and $a|c$ .Let's assume $b=am\,\,; c=an$ .According to the question $a,b,c$ are in A.P. so we can write $2b=a+c \implies 2am=an+a \implies 2m=n+1 \implies n=2m-1$ .Now our new $a,b,c$ becomes $a,am,a(2m-1)$ .Putting this in the given quadratics

$ax^2+amx+a(2m-1)=0 \implies x^2+mx+(2m-1)=0$ .The roots are integer then its\'s discriminant will be a perfect square.

$d^2=m^2-4(2m-1)=m^2-8m+4=(m-4)^2-12\ \implies (m-4)^2-d^2=12 \implies (m-4+d)(m-4-d)=12$

$(m-4+d)(m-4-d)=12=2 \times 6=(-2) \times (-6)$ .To get an integral $m$ the two factors must be even

$m-4+d=6 ; m-4-d=2 \implies m=8$ .The second value will give $m=0$ which is not allowable for non-zero A.P.

Putting $m=8$ the $a,b,c$ becomes $a,8m,15m$ .Now $\alpha+\beta=-8$ and $\alpha \beta=15$

So, $\alpha + \beta=-8+15=\boxed{7}$