Let $$S_n$$ denote the sum of the first n terms of an arithmetic progression. If $$S_{10} = 390$$ and the ratio of the tenth and the fifth terms is 15 : 7, then $$S_{15} - S_5$$ is equal to:

Given that the sequence is an arithmetic progression (AP), let the first term be $$a$$ and the common difference be $$d$$.

The sum of the first n terms is given by $$S_n = \frac{n}{2} [2a + (n-1)d]$$.

The nth term is given by $$T_n = a + (n-1)d$$.

1. $$S_{10} = 390$$

2. The ratio of the tenth term to the fifth term is 15:7, i.e., $$\frac{T_{10}}{T_5} = \frac{15}{7}$$

$$T_{10} = a + 9d$$

$$T_5 = a + 4d$$

$$\frac{a + 9d}{a + 4d} = \frac{15}{7}$$

$$a = \frac{3d}{8}$$ $$-(1)$$

Now, using $$S_{10} = 390$$:

$$S_{10} = \frac{10}{2} [2a + 9d] = 5(2a + 9d) = 390$$

$$2a + 9d = 78$$ $$-(2)$$

Substitute equation (1) into equation (2):

$$2 \left( \frac{3d}{8} \right) + 9d = 78$$

$$d = \frac{312}{39} = 8$$

Substitute d = 8 into equation (1):

$$a = \frac{3 \times 8}{8} = \frac{24}{8} = 3$$

Thus, a = 3 and d = 8.

$$S_{15} = \frac{15}{2} [2a + (15-1)d] = \frac{15}{2} [2(3) + 14(8)]$$

$$= 15 \times 59 = 885$$

$$S_5 = \frac{5}{2} [2a + (5-1)d] = \frac{5}{2} [2(3) + 4(8)]$$

$$= \frac{5}{2} [6 + 32] = \frac{5}{2} \times 38$$

$$= 5 \times 19 = 95$$

$$S_{15} - S_5 = 885 - 95 = 790$$