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 ________.

Updated On: Jun 25, 2025
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Correct Answer: 6993

Solution and Explanation

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.

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Concepts Used:

Arithmetic Progression

Arithmetic Progression (AP) is a mathematical series in which the difference between any two subsequent numbers is a fixed value.

For example, the natural number sequence 1, 2, 3, 4, 5, 6,... is an AP because the difference between two consecutive terms (say 1 and 2) is equal to one (2 -1). Even when dealing with odd and even numbers, the common difference between two consecutive words will be equal to 2.

In simpler words, an arithmetic progression is a collection of integers where each term is resulted by adding a fixed number to the preceding term apart from the first term.

For eg:- 4,6,8,10,12,14,16

We can notice Arithmetic Progression in our day-to-day lives too, for eg:- the number of days in a week, stacking chairs, etc.

Read More: Sum of First N Terms of an AP