The sum of the first n terms of an arithmetic progression is given by S_n = n2 − 8n. Find the least integer n for which the nth term is greater than 130.

The sum of the first $n$ terms of an arithmetic progression is given by $S_n = n^2 - 8n$.
The $n^{th}$ term, $a_n$, can be found using the relationship $a_n = S_n - S_{n-1}$.

First, find $S_{n-1}$ by substituting $(n-1)$ for $n$ in the formula for $S_n$:

$$S_{n-1} = (n-1)^2 - 8(n-1)$$ $$S_{n-1} = (n^2 - 2n + 1) - (8n - 8)$$ $$S_{n-1} = n^2 - 2n + 1 - 8n + 8$$ $$S_{n-1} = n^2 - 10n + 9$$

Now, substitute $S_n$ and $S_{n-1}$ into the formula for $a_n$:

$$a_n = S_n - S_{n-1}$$ $$a_n = (n^2 - 8n) - (n^2 - 10n + 9)$$ $$a_n = n^2 - 8n - n^2 + 10n - 9$$ $$a_n = 2n - 9$$

The problem states that the $n^{th}$ term of the sequence is greater than 130:

$$a_n > 130$$

Substitute the expression for $a_n$:

$$2n - 9 > 130$$

Add 9 to both sides:

$$2n > 130 + 9$$ $$2n > 139$$

Divide by 2:

$$n > \frac{139}{2}$$ $$n > 69.5$$

Since $n$ must be an integer (as it represents the term number) and $n$ must be greater than 69.5, the least possible integer value for $n$ is 70.

The least possible value of $n$ is $\boxed{\text{70}}$.