If coefficients of the Quadratic Equation are in AP find $\alpha+\beta +\alpha\beta$. $ax^2+bx+c=0$ is a quadratic equation with integer roots $\alpha,\beta$ and the positive coefficients are in AP.Find $\alpha+\beta +\alpha\beta$.
My attempt:
$\alpha+\beta +\alpha\beta=\frac {-b}{a}+\frac {c}{a}$ and we know $\frac{a+c}{2}=b$.
But I cannot eliminate the $a,b,c$ terms to get the answer $7$ given.
How to do it?Any quick,short,simple solutions?
 A: First of all we need to that if the roots are integer then $S=\alpha+\beta+\alpha\cdot \beta$ is also integer then
$$S=\frac{c-b}{a}=\frac{b-a}{a}=\frac{b}{a}-1\to a|b$$
and once $2b=a+c$ we get that $a|c$ so we can write the terms $(a,b,c)$ like $(a,a(1+r),a(1+2r))$ and then
$$S=r$$
We also have the discriminant:
$$\Delta=b^2-4ac=a^2(1+r)^2-4a\cdot a(1+2r)=a^2[(1+r)^2-4(1+2r)]$$
and once the roots are integer then $\Delta$ should be a square and then
$$(1+r)^2-4(1+2r)=k^2\to r^2-6r-(3+k^2)=0\quad (1)$$
and looking this new quadratic equation (in $r$) we have the discriminant 
$$\Delta '=36-4[-(3+k^2)]=4(12+k^2)$$
and $\Delta'$ has also to be an integer, then:
$$12+k^2=p^2\to (p-k)(p+k)=12$$
spliting $12$ as a product of two numbers we get the solutions $(k,p)\in\{(2,4),(-2,-4),(2,-4),(-2,4)\}$. So, $k=\pm 2\to k^2=4$ and then backing to $(1)$ we get $r=7$ or $r=-1$. 
For $r=-1$ we get the sequence $(a,0,a)$ and the equation $ax^2+a=a(x^2+1)$ what not give us real solution.
For $r=7$ we get the equation $ax^2+8ax+15a=0$ what give us the roots $-3$ and $-5$, then
$$S=r=7$$ 
A: If the given quadratic has integer roots $\alpha,\beta$, then 
$$\begin{align}
ax^2+bx+c
&=a\left(x^2+\frac bax+\frac ca\right)\\
&=a(x-\alpha)(x-\beta)\\
&=a[x^2-(\alpha+\beta)x+\alpha\beta]=0\\
\begin{cases}
\frac ba=-(\alpha+\beta)\\\frac ca=\alpha\beta\end{cases}\end{align}$$
Given that $a,b,c$ are in AP, then $1, \frac ba, \frac ca$ are also in AP.
The problem reduces to finding $\alpha+\beta+\alpha\beta$ where $1, -(\alpha+\beta), \alpha\beta$ are positive integers in AP. 
For the three numbers to be positive, $\alpha, \beta<0 \; (\alpha, \beta\in\mathbb Z)$. 
For them to be in AP, 
$$\begin{align}
-2(\alpha+\beta)&=1+\alpha\beta\\
\beta&=-2+\frac 3{\alpha+2}
\end{align}$$
Since $\beta$ is an integer, $\frac 3{\alpha+2}$must be an integer.
As $\alpha,\beta <0$ and $\alpha, \beta\in\mathbb Z$, possible values for $\alpha, \beta$ are $(-3,-5), (-5,-3)$.
Hence $$\alpha+\beta+\alpha\beta=\color{red}7$$
