The quadratic polynomial $ax^2+bx+c$ has positive coefficients $a,b,c$ in A.P. in the given order. If it has integer roots $\alpha,\beta,$ find $\alpha+\beta+\alpha \beta.$
I tried with Vieta's theorem and putting $b=\frac{a+c}{2}$ to get $\alpha+\beta+\alpha \beta=\frac{b}{a}-1=\frac{c-a}{2a}$ but couldn't arrive at a solution.
P.S. The question had the following options given of which one and only one is the correct answer (if they are of any help)-$3,5,7,14$.
The roots have sum $-\frac{b}{a}$ and product $\frac{c}{a}$, and these sum to $\frac{c-b}{a}=\frac{b}{a}-1$. Comparing this to the sum of the roots, $(\alpha +2)(\beta +2)=3$. As the roots are integers, $\alpha, \,\beta $ are $-1,\,1$ or $-5,\,-3$ in some order. These both give a suitable quadratic, with the desired quantity either $-1$ or $7$.
Unpacking J.G.'s answer for the masses:
$ax^2+bx+c$
$$\text{if} \quad (a,b,c) \quad \text{are in "AP", then} \quad $$ $$\begin{align}\begin{cases} a&=a \\ b&=a+d \\ c&=a+2d \end{cases} \ \ &\underbrace{\implies}_{\text{b is the average of a and c}} \ \ \left[\frac{a+c}{2} =\frac{2a+2d}{2}=a+d=b\right] \\ &\qquad \quad \implies \qquad \quad \ \left[c=2b-a\right]\end{align}$$
Vieta: if $\alpha$ and $\beta$ are the roots $\implies \frac{-b}{a}=\alpha + \beta$ and $\frac{c}{a}=\frac{2b-a}{a}=\alpha \beta $
$$\begin{align} \implies \alpha+\beta+\alpha \beta&=\frac{c-b}{a} \\ &=\frac{b-a}{a} \\ &=\frac{b}{a}-1 \\ &=-(\alpha+\beta)-1 \\ \alpha \beta +2(\alpha+\beta)&=-1 \\ \alpha \beta +2(\alpha +\beta)+4&=3 \\ (\alpha+2)(\beta+2)&=3=d_1 \cdot d_2 \\ \text{such that}\quad (d_1,d_2) &\in \{(1,3),(3,1),(-1,-3),(-3,-1) \\ &\begin{cases} \alpha=d_1-2 \\ \beta=d_2-2 \end{cases} \\ \text{so} \quad (\alpha,\beta)&\in\{(-1,1),(1,-1),(-3,-5),(-5,-3) \\ \implies \alpha + \beta + \alpha \beta&=-1 \quad \lor \quad 7 \end{align}$$
