Let $$s_1, s_2, s_3, \ldots, s_{10}$$ respectively be the sum of 12 terms of 10 A.P.s whose first terms are $$1, 2, 3, \ldots, 10$$ and the common differences are $$1, 3, 5, \ldots, 19$$ respectively. Then $$\sum_{i=1}^{10} s_i$$ is equal to

We have 10 arithmetic progressions where the $$i$$-th AP has first term $$a_i = i$$ and common difference $$d_i = 2i - 1$$ for $$i = 1, 2, \ldots, 10$$.

The sum of 12 terms of the $$i$$-th AP is:

$$s_i = \frac{12}{2}\left(2a_i + 11d_i\right) = 6\left(2i + 11(2i-1)\right) = 6(2i + 22i - 11) = 6(24i - 11)$$

Now we compute the total:

$$\sum_{i=1}^{10} s_i = 6 \sum_{i=1}^{10}(24i - 11) = 6\left(24 \cdot \frac{10 \times 11}{2} - 11 \times 10\right)$$

$$= 6\left(24 \times 55 - 110\right) = 6\left(1320 - 110\right) = 6 \times 1210 = 7260$$

The correct answer is Option C: $$7260$$.