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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

S 11=3[101+102+ ldots ldots+121] =(3/2)(222) × 21=6993

Q. The series of positive multiples of $3$ is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21$ , $24, 27}, \dots$ Then the sum of the elements in the $1 1^{th}$ set is equal to_______

$S_{11} = 3 [101 + 102 + \dots\dots + 121]$
$= \frac{3}{2} (222) \times 21 = 6993$

Q. 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 11th set is equal to_______

S11​=3[101+102+……+121] =23​(222)×21=6993

Q. The series of positive multiples of $3$ is divided into sets : ${3}, {6, 9, 12}, {15, 18, 21$ , $24, 27}, \dots$ Then the sum of the elements in the $1 1^{th}$ set is equal to_______

$S_{11} = 3 [101 + 102 + \dots\dots + 121]$
$= \frac{3}{2} (222) \times 21 = 6993$