The series of positive multiples of 3 is divided into sets {3}, {6,9, 12}, {15, 19,21,24,27},....... Then the sum of
the elements in the 11 th set is equal to________

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answer is 6993.

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Detailed Solution

$\begin{array}{l} ∴ 11^{th} set will have 1 + (10) 2 = 21 terms \\ Number of elements upto 10 th set = 1 + 3 + 5 + \dots \dots + 19 = 100 terms \\ S_{11} = 3 [101 + 102 + \dots \dots + 121] \\ = \frac{3}{2} (222) \times 21 = 6993 \end{array}$

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