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.
Recall that the explicit formula for an arithmetic series is:
$$a_n = a_0 + nd \tag{1}$$
... where $d$ is the common difference.
So, you're given that $a_{9} = 26$ and that $a_0 = 5$. With that information, can you solve equation $(1)$ for $d$?
For the sum question, what you want is the first $n$ such that:
$$\sum_{k=0}^{n}a_k \gt 1000 \tag{2}$$
This can be done brute-force with a calculator (or pencil and paper with a lot of patience), or with formulas.
We know that $\sum_{k=0}^{n} k = \frac{n(n+1)}{2}$, and that $\sum_{k=0}^{n}1 = n+1$. Thus, plugging equation $(1)$ in to equation $(2)$:
$$\sum_{k=0}^{n} a_0 + kd \gt 1000 $$
$$\sum_{k=0}^{n} a_0 + \sum_{k=0}^{n}kd \gt 1000$$
$$a_0\sum_{k=0}^{n} 1 + d\sum_{k=0}^{n}k \gt 1000$$
$$a_0(n+1) + d\frac{n(n+1)}{2}\gt 1000\tag{3}$$
So, all you need to do is solve $(3)$ for the smallest integer $n$ for which the equality holds.
Best Answer
HINT:
As the number of terms is odd, let the middle $\left(\frac{5+1}2=3\text{rd}\right)$ term be $a$
So, the $5$ terms will be $a,a\pm d,a\pm 2d$
So, $(a-2d)+(a-d)+a+(a+d)+(a+2d)=30\implies 5a=30\implies a=6$
and $(a-2d)^2+(a-d)^2+a^2+(a+d)^2+(a+2d)^2=220$
$\implies a^2+2\{a^2+d^2\}+2\{a^2+(2d)^2\}=220$ as $(A+B)^2+(A-B)^2=2(A^2+B^2)$
$\implies 10d^2=220-5\cdot 6^2=40\implies d^2=4$
Had the number of terms been even, we could take the terms as $a\pm d,a\pm3d,\cdots$