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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 \(................\)

  1. A \(6994\)
  2. B \(6698\)
  3. C \(6695\)
  4. D \(6993\)
Verified Solution

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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