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 $$T_n = a + (n-1)d$$.

Given conditions:

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

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

First, express the terms:

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

$$T_5 = a + 4d$$

Using the ratio:

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

Cross-multiplying:

$$7(a + 9d) = 15(a + 4d)$$

$$7a + 63d = 15a + 60d$$

$$7a + 63d - 15a - 60d = 0$$

$$-8a + 3d = 0$$

$$3d = 8a$$

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

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):

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

$$\frac{6d}{8} + 9d = 78$$

$$\frac{3d}{4} + 9d = 78$$

Convert to common denominator:

$$\frac{3d}{4} + \frac{36d}{4} = 78$$

$$\frac{39d}{4} = 78$$

Multiply both sides by 4:

$$39d = 312$$

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

Substitute d = 8 into equation (1):

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

Thus, a = 3 and d = 8.

Now, compute $$S_{15} - S_5$$:

First, find $$S_{15}$$:

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

$$= \frac{15}{2} [6 + 112] = \frac{15}{2} \times 118$$

$$= 15 \times 59 = 885$$

Next, find $$S_5$$:

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

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

$$= 5 \times 19 = 95$$

Therefore,

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

Alternatively, $$S_{15} - S_5$$ is the sum of terms from the 6th to the 15th term (10 terms).

The 6th term: $$T_6 = a + 5d = 3 + 5 \times 8 = 43$$

The 15th term: $$T_{15} = a + 14d = 3 + 14 \times 8 = 115$$

Sum = $$\frac{10}{2} (43 + 115) = 5 \times 158 = 790$$

Hence, the answer is 790, which corresponds to option C.