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

Given,

First term of A.P are $1,2,3....10$, so general term of the first term will be $i$

And common difference are $1,3,5,....$, so general term of common difference is given by $2i-1$

Now sum of the A.P  is given by,

$S_i=\frac{12}{2}\left[2\times i+\left(12-1\right)\left(2i-1\right)\right]$

$\Rightarrow S_i=6\left[2\times i+11\left(2i-1\right)\right]$

$\Rightarrow S_i=144i-66$

So, $\sum _{i=1}^{10}{S}_i=\sum _{i=1}^{10}144i-66\sum _{i=1}^{10}1$

$\Rightarrow \sum _{i=1}^{10}{S}_i=144\left(\frac{10\times 11}{2}\right)-66\times 10$

$\Rightarrow \sum _{i=1}^{10}{S}_i=792\left(10\right)-66\times 10$

$\Rightarrow \sum _{i=1}^{10}{S}_i=7260$

Hence this is the correct option.