Question

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 ________.

Answer 1

Given series
∴ 11th set will have 1 + (10)2 = 21 term
Also upto 10th set total 3 × k type terms will be
1 + 3 + 5 +……+19 = 100 – term
∴ Set 11 = {3 × 101, 3 × 102,……3 × 121}
∴ Sum of elements = 3 × (101 + 102 +…+121)
$=\frac{3×222×21}{2}=6993$
So, the correct option is 6993.