The $$8^{th}$$ common term of the series
$$S_1 = 3 + 7 + 11 + 15 + 19 + \ldots$$
$$S_2 = 1 + 6 + 11 + 16 + 21 + \ldots$$ is

We need to find the 8th common term of the series $$S_1 = 3 + 7 + 11 + 15 + 19 + \ldots$$ and $$S_2 = 1 + 6 + 11 + 16 + 21 + \ldots$$

For $$S_1$$, the first term is $$3$$ and the common difference is $$4$$, so the general term is $$3 + (n-1)\cdot4 = 4n - 1$$.

For $$S_2$$, the first term is $$1$$ and the common difference is $$5$$, so the general term is $$1 + (m-1)\cdot5 = 5m - 4$$.

Setting $$4n - 1 = 5m - 4$$ gives $$4n = 5m - 3$$.

The first common term is $$11$$ (when $$n = 3, m = 3$$).

The common difference of the common terms is $$\text{lcm}(4, 5) = 20$$.

The common terms form an A.P.: $$11, 31, 51, 71, 91, 111, 131, 151, \ldots$$

The general term of this A.P. is $$11 + (k - 1)\times20 = 20k - 9$$.

For the 8th term, $$a_8 = 11 + 7 \times 20 = 11 + 140 = 151$$.

The answer is $$\boxed{151}$$.