JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
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 ________.
Sequence and Series · Arithmetic Progression
Answer$6993$
Solution
Formulas used. (1) $1+3+5+\cdots+(2k-1)=k^2$(2) Sum of an A.P.: $\frac n2\,(\text{first}+\text{last})$Step 1 - Set sizes. The sets have $1,3,5,\dots$ elements; the $11$th set has $21$. Step 2 - Where the $11$th set starts. The first $10$ sets hold $10^2=100$ multiples of $3$. So the $11$th set is $$3\times101,\ 3\times102,\ \dots,\ 3\times121 .$$ Step 3 - Sum. $$3\left(101+102+\cdots+121\right)=3\cdot\frac{21}{2}(101+121)=3\times2331=6993 .$$ Answer: 6993