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

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Answer: C. 7260 We have 10 arithmetic progressions (A.P.s) with the first terms a_i and the common differences d_i, where i = 1, 2, \ldots, 10. The first terms are a_i = i and the common differences are d_i = 2i - 1. Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first n terms of an A.P. is: S_n = n\left(\frac{2a + (n - 1)d}{2}\right) In this case, we need to find the sum of the first 12 terms for each A.P., so we have: S_{12} = 12\left(\frac{2a + 11d}{2}\right) Now, we can compute the sum s_i for each A.P.: s_i = 12\left(\frac{2i + 11(2i - 1)}{2}\right) = 6(2i + 22i - 11) = 6(24i - 11) Finally, we need to find the sum of all s_i for i = 1, 2, \ldots, 10: \sum\limits_{i=1}^{10} s_i = 6\sum\limits_{i=1}^{10} (24i - 11) = 6\left(24\sum\limits_{i=1}^{10} i - 11\sum\limits_{i=1}^{10} 1\right) The sum of the first 10 integers is \sum\limits_{i=1}^{10} i = \frac{10(10 + 1)}{2} = 55, so we have: \sum\limits_{i=1}^{10} s_i = 6\left(24\cdot55 - 11\cdot10\right) = 6(1320 - 110) = 6\cdot1210 = 7260 Thus, the sum \sum\limits_{i=1}^{10} s_i is equal to 7260.

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

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