For the $i$-th A.P.:

• First term $a_i = i$, where $i=1,2,\dots,10$
• Common difference $d_i = 2i-1$ since the differences are $1,3,5,\dots,19$

Sum of first $12$ terms of an A.P. is $s_i=\frac{12}{2}\left[2a_i+(12-1)d_i\right]$ $s_i=6\left[2i+11(2i-1)\right]$

Simplify: $s_i=6\left[2i+22i-11\right]$ $s_i=6(24i-11)$ $s_i=144i-66$

Now sum from $i=1$ to $10$: $\sum_{i=1}^{10} s_i = \sum_{i=1}^{10} (144i-66)$ $=144\sum_{i=1}^{10} i - 66\cdot 10$

Use $\sum_{i=1}^{10} i = \frac{10\cdot 11}{2}=55$ Therefore, $\sum_{i=1}^{10} s_i = 144\cdot 55 - 660$ $=7920-660$ $=7260$

Hence the correct option is $\boxed{7260}$ which is option $\boxed{\text{C}}$.