JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Sequences and Series

The $n$-th set contains $2n-1$ consecutive multiples of 3. The first set has 1 element, the second has 3, the third has 5, etc. The total number of elements in the first $n$ sets is $1+3+5+\cdots +(2n-1)=n^2$. Thus the first 10 sets contain $10^2=100$ elements, and the 11th set contains elements from position 101 to 121 (21 elements). These are the 101st through 121st multiples of 3: $3\cdot 101,3\cdot 102,\dots ,3\cdot 121$. The sum is $3(101+102+\cdots +121)=3\cdot \frac{21(101+121)}{2}=3\cdot \frac{21\cdot 222}{2}=3\cdot 21\cdot 111=6993$.