Let $$S_n$$ denote the sum of first $$n$$-terms of an arithmetic progression. If $$S_{10} = 530$$, $$S_5 = 140$$, then $$S_{20} - S_6$$ is equal to:

For an arithmetic progression with first term $$a$$ and common difference $$d$$, the sum of the first $$n$$ terms is $$S_n = \frac{n}{2}[2a + (n-1)d]$$.

Given $$S_5 = 140$$: $$\frac{5}{2}[2a + 4d] = 140$$, so $$2a + 4d = 56$$, giving $$a + 2d = 28$$ $$-(1)$$.

Given $$S_{10} = 530$$: $$\frac{10}{2}[2a + 9d] = 530$$, so $$2a + 9d = 106$$ $$-(2)$$.

From $$(1)$$: $$2a = 56 - 4d$$. Substituting into $$(2)$$: $$56 - 4d + 9d = 106$$, so $$5d = 50$$, giving $$d = 10$$.

From $$(1)$$: $$a = 28 - 2(10) = 8$$.

Now we compute $$S_{20}$$: $$S_{20} = \frac{20}{2}[2(8) + 19(10)] = 10[16 + 190] = 10 \times 206 = 2060$$.

And $$S_6$$: $$S_6 = \frac{6}{2}[2(8) + 5(10)] = 3[16 + 50] = 3 \times 66 = 198$$.

Therefore, $$S_{20} - S_6 = 2060 - 198 = 1862$$.

The answer is $$1862$$, which is Option A.