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 s_i is equal to

(c) : s_1=1+2+3+ * s+12 s_2=2+5+8+ * s upto 12 terms\ s_3=3+8+13+ * s upto 12 terms \ Because s_10=10+29+48+ * s upto 12 terms From (i), s_1=(12(13) / 2)=78 From (ii), s_2=(12 / 2)[2(2)+11 x 3] =6[4+33]=222 From (iii), s_3=(12 / 2)[2(3)+11 x 5] =6[6+55]=366 From (iv), s_10=(12 / 2)[2(10)+11 x 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((10 x 11 / 2))-66(10) =7920-660=7260

Let s_{1},s_{2},s_{3},..s_{10} respectively be the sum ... | 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

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