This is what I'd write:
all((i - j) == (j - k) for i, j, k in zip(l[:-2], l[1:-1], l[2:]))
You could probably make it more efficient by only computing the differences once, but if you're concerned about efficiency you'd use numpy and write:
np.all((a[:-2] - a[1:-1]) == (a[1:-1] - a[2:]))
or even (saving a slice):
np.all(a[:-2] + a[2:] == 2 * a[1:-1])
Probably the most concise method is to use numpy.diff, as it will automatically convert a list into a numpy array:
np.all(np.diff(l, 2) == 0)