Let s_1,s_2,,s_10 respectively be the sums to 12 terms of 10 A.P.s whose first terms are 1…

Question

Let $s_1,s_2,\ldots,s_{10}$ respectively be the sums to $12$ terms of $10$ A.P.s whose first terms are $1,2,3,\ldots,10$ and common differences are $1,3,5,\ldots,19$ respectively. Then $\displaystyle\sum_{i=1}^{10}s_i$ is equal to:

Accepted Answer

$s_k = \dfrac{12}{2}\big[2k+11(2k-1)\big] = 6(24k-11) = 144k-66.$

$$\sum_{k=1}^{10}s_k = 144\sum_{k=1}^{10}k - 660 = 144(55)-660 = 7920-660 = 7260.$$

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