To prove an Arithmetic Progression If the $p^{th}$ term of an arithmetic progression is $\alpha$ and the $q^{th}$ term is $\beta$, prove that the sum of its $p+q$ term is 
$$\begin{align}\frac{p+q}{2}\frac{(\alpha+\beta)+(\alpha-\beta)}{p-q}\end{align}.$$
Here can we take the $p^{th}$ and $q^{th}$ term as consecutive terms of the arithmetic progression?
 A: 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.
A: If $a,d$ be the 1st term & the common difference respectively,
$\alpha=a+(p-1)d$ and $\beta=a+(q-1)d$
Solve for $a,d$
The sum will be $\frac{p+q}2\{2a+(p+q-1)d\}$
A: $$x_p=x_1+(p-1)d=\alpha$$
$$x_q=x_1+(q-1)d=\beta$$
$$x_1+(p-1)d-(x_1+(q-1)d)=\alpha-\beta$$
$$pd-d-qd+d=\alpha-\beta$$
$$d=\frac{\alpha-\beta}{p-q}$$
$$x_1=\alpha-(p-1)d$$
$$S_{p+q}=\frac{p+q}{2}(2x_1+(p+q-1)d)$$
$$S_{p+q}=\frac{p+q}{2}\left(2(\alpha-(p-1)\frac{\alpha-\beta}{p-q})+(p+q-1)\frac{\alpha-\beta}{p-q}\right)$$
$$S_{p+q}=\frac{p+q}{2}\left(\frac{\alpha-\beta}{p-q}(2\alpha-p+q+1)\right)$$
