JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(7260\)
Answer & Solution
Correct Answer
(D) \(7260\)
Step-by-step Solution
Detailed explanation
\(S _{ k }=6(2 k +(11)(2 k -1))\) \(S _{ k }=6(2 k +22 k -11)\) \(S _{ k }=144 k -66\) \(\sum \limits_1^{10} S _{ k }=144 \sum \limits_{ k =1}^{10} k -66 \times 10\)
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