Let s_1, s_2, s_3, ldots, s_{10} be the sum o | vedclass

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(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.For the $k$-th arithmetic progression,th

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$7260$

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(D) The sum of the first $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.For the $k$-th arithmetic progression,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.Given $n = 12$,the sum $s_k$ is:$s_k = \frac{12}{2} [2(k) + (12-1)(2k-1)]$$s_k = 6 [2k + 11(2k-1)]$$s_k = 6 [2k + 22k - 11] = 6 [24k - 11] = 144k - 66$.Now,we calculate the sum $\sum_{i=1}^{10} s_i$:$\sum_{i=1}^{10} (144i - 66) = 144 \sum_{i=1}^{10} i - \sum_{i=1}^{10} 66$$= 144 	imes \frac{10 	imes 11}{2} - 660$$= 144 	imes 55 - 660$$= 7920 - 660 = 7260$.

Let $s_1, s_2, s_3, \ldots, s_{10}$ be the sum of the first $12$ terms of $10$ arithmetic progressions whose first terms are $1, 2, 3, \ldots, 10$ and whose common differences are $1, 3, 5, \ldots, 19$ respectively. Then $\sum_{i=1}^{10} s_i$ is equal to

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