We have 10 arithmetic progressions (A.P.s) with the first terms $$a_i$$ and the common differences $$d_i$$, where $$i = 1, 2, \ldots, 10$$.

The first terms are $$a_i = i$$ and the common differences are $$d_i = 2i - 1$$.

Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first n terms of an A.P. is:

$$S_n = n\left(\frac{2a + (n - 1)d}{2}\right)$$

In this case, we need to find the sum of the first 12 terms for each A.P., so we have:

$$S_{12} = 12\left(\frac{2a + 11d}{2}\right)$$

Now, we can compute the sum $$s_i$$ for each A.P.:

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

Finally, we need to find the sum of all $$s_i$$ for $$i = 1, 2, \ldots, 10$$:

$$\sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right)$$

The sum of the first 10 integers is $$\sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55$$, so we have:

$$\sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260$$

Thus, the sum $$\sum\limits_{i=1}^{10} s_i$$ is equal to 7260.