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 _______.

Correct answer: 0 Given, \ 3\ , \ 6, 9, 12\ , \ 15, 18, 21, 24, 27\ , ⋯ Now, the number of elements in 11 ^ th set will be = 1 + (10)2 = 21 The total number of elements up to 10 ^ th set will be 1 + 3 + . . . + 19= 10 ^ 2 = 100 ∴ elements in 11 ^ th set = \ 3 · 101, 3 · 102, . . . . 3 · 121\ Sum of these elements = 3 (101 + 102 + ⋯ + 121) = 3 × (21)/(2) × (101 + 121) = 6993

JEE Main 202226 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 26 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 26 July · Morning Shift · Mathematics · Sequences and Series

The series of positive multiples of 3 is divided into sets : \ 3\ ,...

The series of positive multiples of

3

is divided into sets : \left\{3\right\}, \left\{6, 9, 12\right\}, \left\{15, 18, 21, 24, 27\right\}, \ldots Then the sum of the elements in the

{11}^{th}

set is equal to _______.

Correct answer

Step-by-step solution

Given, $\left\{3\right\}, \left\{6, 9, 12\right\}, \left\{15, 18, 21, 24, 27\right\}, ⋯$
Now, the number of elements in ${11}^{\text{th}}$ set will be $= 1 + \left(10\right)2 = 21$
The total number of elements up to ${10}^{\text{th}}$ set will be $1 + 3 + . . . + 19= {10}^{2} = 100$
$∴$ elements in ${11}^{\text{th}}$ set $= \left\{3 \cdot 101, 3 \cdot 102, . . . . 3 \cdot 121\right\}$
Sum of these elements $= 3 \left(101 + 102 + ⋯ + 121\right)</p><p>$ = 3 \times $\dfrac{21}{2}$ \times \left(101 + 121\right) $= 6993$

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