The series of positive multiples of 3 is divi | vedclass

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(D) The number of elements in the $n^{	ext{th}}$ set is $2n - 1$.The total number of elements in the first $10$ sets is $\sum_{k=1}^{10} (2k - 1) = 2

Vedclass · Accepted Answer

$6993$

Vedclass · Answer

(D) The number of elements in the $n^{	ext{th}}$ set is $2n - 1$.The total number of elements in the first $10$ sets is $\sum_{k=1}^{10} (2k - 1) = 2 	imes \frac{10 	imes 11}{2} - 10 = 110 - 10 = 100$.Thus,the $11^{	ext{th}}$ set starts from the $101^{	ext{st}}$ multiple of $3$,which is $3 	imes 101 = 303$.The $11^{	ext{th}}$ set contains $2(11) - 1 = 21$ elements.These elements form an arithmetic progression with first term $a = 303$,common difference $d = 3$,and number of terms $n = 21$.The sum of the elements is $S_{11} = \frac{n}{2} [2a + (n - 1)d] = \frac{21}{2} [2(303) + (21 - 1)3] = \frac{21}{2} [606 + 60] = \frac{21}{2} [666] = 21 	imes 333 = 6993$.

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 $................$

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