The series of positive multiples of 3 is divided into sets: \{ 3\},\{ 6,9,12\},\{ 15,18,21,24,27\},.. Then the sum of the elements in the 11^{{th}{~}} set is equal to: Sequence & Series (Mathematics)

Let $a_i$ denotes the $i^{\text {th }}$ set. $\\$ ∴ Number of elements in $a_1=1\\$ Number of elements in $a_2=3\\$ Number of elements in $a_3=5\\$ ∴ Number of elements in $a_{11}=1+(11-1) \cdot 2=21\\$ Total number of elements in set $1+$ set $2+\cdots+$ set 10 $\\=1+3+5+\cdots+1+(9 \times 2)=100\\$ ∴ First term of $a_{11}=101 \times 3=303\\$ and last term of $a_{11}=121 \times 3=363\\$ ∴ Sum of elements of $a_{11}=\frac{21}{2}(303+363)=6993$

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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