Let s_{1},s_{2},s_{3},..s_{10} respectively be the sum to 12 terms of 10 A.P. whose first terms are 1,2,3,. 10 and the common difference are 1,3,5,..,19 respectively. Then \sum_{i = 1}^{10}\mspace{2mu} s_{i} is equal to: Sequence & Series (Mathematics)

(c) : $s_1=1+2+3+\cdots+12\\$ $s_2=2+5+8+\cdots$ upto 12 terms $\\ s_3=3+8+13+\cdots$ upto 12 terms $\\ \because s_{10}=10+29+48+\cdots \\$ upto 12 terms From (i), $s_1=\frac{12(13)}{2}=78\\$ From (ii), $s_2=\frac{12}{2}[2(2)+11 \times 3] \\=6[4+33]=222\\$ From (iii), $s_3=\frac{12}{2}[2(3)+11 \times 5] \\=6[6+55]=366\\$ From (iv), $s_{10}=\frac{12}{2}[2(10)+11 \times 19] \\=6[20+209]=1374\\$ Thus, $s_k=6(2 k+11(2 k-1)) \\=6(2 k+22 k-11)=144 k-66\\$ $\therefore \sum_{k=1}^{10} s_k\\=144 \sum_{k=1}^{10} k-66 \sum_{k=1}^{10} 1\\=144\left(\frac{10 \times 11}{2}\right)-66(10)\\$ $=7920-660=7260$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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