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

We have 10 arithmetic progressions (A.P.s) with the first terms $a_i$ and the common differences $d_i$, where $i = 1, 2, \ldots, 10$. <br/><br/>The first terms are $a_i = i$ and the common differences are $d_i = 2i - 1$. <br/><br/>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: <br/><br/>$S_n = n\left(\frac{2a + (n - 1)d}{2}\right)$ <br/><br/>In this case, we need to find the sum of the first 12 terms for each A.P., so we have: <br/><br/>$S_{12} = 12\left(\frac{2a + 11d}{2}\right)$ <br/><br/>Now, we can compute the sum $s_i$ for each A.P.: <br/><br/>$s_i = 12\left(\frac{2i + 11(2i - 1)}{2}\right) = 6(2i + 22i - 11) = 6(24i - 11)$ <br/><br/>Finally, we need to find the sum of all $s_i$ for $i = 1, 2, \ldots, 10$: <br/><br/>$$\sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right)$$ <br/><br/>The sum of the first 10 integers is $\sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55$, so we have: <br/><br/>$$\sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260$$ <br/><br/>Thus, the sum $\sum\limits_{i=1}^{10} s_i$ is equal to 7260.