Let the arithmetic progression be $\langle x_1,x_2,x_3,\dots\rangle$, so that $\alpha=x_p$ and $\beta=x_q$. Let $S=x_1+x_2+\ldots+x_{p+q}\;,$ the sum of the first $p+q$ terms. Let $d$ be the constant difference of this progression, so that $x_{k+1}=x_k+d$ for every $k$. Then $x_{q+1}=x_q+d$, $x_{q+2}=x_q+2d$, and in general $x_{q+k}=x_q+kd$. In particular, if we set $k=p-q$, then $x_p=x_{q+k}=x_q+kd$, i.e., $\alpha=\beta+(p-q)d$, and it follows that $d=\frac{\alpha-\beta}{p-q}\;.$
Now write out the sum $S$ twice, as shown below, and add:
$\begin{array}{c} S&=&x_1&+&x_2&+&\ldots&+&x_{p+q-1}&+&x_{p+q}\\ S&=&x_{p+q}&+&x_{p+q-1}&+&\ldots&+&x_2&+&x_1\\ \hline \end{array}$
On the left you get $2S$. Each column on the right has the form: it contains $x_k$ and $x_{p+q+1-k}$, and its sum is $x_k+x_{p+q+1-k}$. In particular, for $k=p$ we get the sum $x_p+x_{q+1}=\alpha+\beta+d$.
Each column has the same sum (why?), and there are $p+q$ columns, so
$2S=(p+q)(\alpha+\beta+d)\;.$
If you now put all of the pieces together properly, you’ll get the desired formula.