Let s_1,s_2,s_3....,s_10 respectively be the sum of 12 terms of 10A. Ps whose first terms are 1,2,3,....,10 and the common differences are…

Given,

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

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

Now sum of the A.P is given by,

$${S}_{i}=\frac{12}{2}[2\times i+(12-1)(2i-1)]$$Si​=212​[2×i+(12−1)(2i−1)]

$$\Rightarrow {S}_{i}=6[2\times i+11(2i-1)]$$⇒Si​=6[2×i+11(2i−1)]

$$\Rightarrow {S}_{i}=144i-66$$⇒Si​=144i−66

So, $\displaystyle \sum _{i=1}^{10}{S}_{i}=\sum _{i=1}^{10}144i-66\sum _{i=1}^{10}1$i=1∑10​Si​=i=1∑10​144i−66i=1∑10​1

$$\Rightarrow \sum _{i=1}^{10}{S}_{i}=144(\frac{10\times 11}{2})-66\times 10$$⇒i=1∑10​Si​=144(210×11​)−66×10

$$\Rightarrow \sum _{i=1}^{10}{S}_{i}=792(10)-66\times 10$$⇒i=1∑10​Si​=792(10)−66×10

$$\Rightarrow \sum _{i=1}^{10}{S}_{i}=7260$$⇒i=1∑10​Si​=7260

Hence this is the correct option.