Let $$S_a$$ denote the sum of first $$n$$ terms an arithmetic progression. If $$S_{20} = 790$$ and $$S_{10} = 145$$, then $$S_{15} - S_5$$ is :

Let $$S_n$$ denote the sum of first $$n$$ terms of an AP with first term $$a$$ and common difference $$d$$.

The formula is: $$S_n = \frac{n}{2}[2a + (n-1)d]$$

Given: $$S_{20} = 790$$ and $$S_{10} = 145$$.

$$ S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d) = 790 $$

$$ \Rightarrow 2a + 19d = 79 \quad ...(1) $$

$$ S_{10} = \frac{10}{2}[2a + 9d] = 5(2a + 9d) = 145 $$

$$ \Rightarrow 2a + 9d = 29 \quad ...(2) $$

Subtracting (2) from (1): $$10d = 50 \Rightarrow d = 5$$.

From (2): $$2a + 45 = 29 \Rightarrow 2a = -16 \Rightarrow a = -8$$.

Now computing $$S_{15}$$ and $$S_5$$:

$$ S_{15} = \frac{15}{2}[2(-8) + 14(5)] = \frac{15}{2}[-16 + 70] = \frac{15}{2}(54) = 405 $$

$$ S_5 = \frac{5}{2}[2(-8) + 4(5)] = \frac{5}{2}[-16 + 20] = \frac{5}{2}(4) = 10 $$

$$ S_{15} - S_5 = 405 - 10 = 395 $$

The answer is Option (1): $$\boxed{395}$$.