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

Let a_i denotes the i^ 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) * 2=21 Total number of elements in set 1+ set 2+ * s+ set 10 =1+3+5+ * s+1+(9 x 2)=100 ∴ First term of a_11=101 x 3=303 and last term of a_11=121 x 3=363 ∴ Sum of elements of a_11=(21 / 2)(303+363)=6993

The series of positive multiples of 3 is divided into s... | 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

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