JEE Mains · Maths · STD 11 - 8. sequence and series
The series of positive multiples of \(3\) is divided into sets : \(\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots\) Then the sum of the elements in the \(11^{\text {th }}\) set is equal to \(................\)
- A \(6994\)
- B \(6698\)
- C \(6695\)
- D \(6993\)
Answer & Solution
Correct Answer
(D) \(6993\)
Step-by-step Solution
Detailed explanation
\(S _{11}=3[101+102+\ldots \ldots+121]\) \(=\frac{3}{2}(222) \times 21=6993\)
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