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8. Sequences and Series
hard
Let $s _1, s _2, s _3, \ldots \ldots, s _{10}$ 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 $1,3,5, \ldots, 19$ respectively. Then $\sum \limits_{i=1}^{10} s _{ i }$ is equal to
A
$7380$
B
$7220$
C
$7360$
D
$7260$
(JEE MAIN-2023)
Solution
$S _{ k }=6(2 k +(11)(2 k -1))$
$S _{ k }=6(2 k +22 k -11)$
$S _{ k }=144 k -66$
$\sum \limits_1^{10} S _{ k }=144 \sum \limits_{ k =1}^{10} k -66 \times 10$
Standard 11
Mathematics
