Let s1, s2, s3,…..,s10 respectively be the sum to 12 terms of 10 A.P.s whose first terms are 1,2,3, …. ,10 and the common differences are 1, 3, 5, …. , 19 respectively. Then \(10∑^{10}_{i=1} s_i\) is
- 7260
- 7220
- 7360
- 7380
The Correct Option is A
Solution and Explanation
Denote by \(s_k\) the sum of 12 terms of the \(k\)-th arithmetic progression, whose first term is \(k\) and common difference is \(2k-1\). We recall that the sum of 12 terms of an A.P. with first term \(a\) and common difference \(d\) is \[ S_{12} = \frac{12}{2} [2a + (12 - 1)d] = 6(2a + 11d). \] Hence for the \(k\)-th progression, \[ s_k = 6(2k + 11(2k - 1)) = 6(2k + 22k - 11) = 6(24k - 11) = 144k - 66. \] We wish to find \[ \sum_{k=1}^{10} s_k = \sum_{k=1}^{10} (144k - 66) = 144 \sum_{k=1}^{10} k - 66 \sum_{k=1}^{10} 1. \] Recall \[ \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55, \] and \[ \sum_{k=1}^{10} 1 = 10. \] Thus \[ \sum_{k=1}^{10} s_k = 144 \cdot 55 - 66 \cdot 10 = 7920 - 660 = 7260. \] Hence, \[ \boxed{ \sum_{k=1}^{10} s_k = 7260. } \]
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