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Q. The natural numbers are grouped as follows:
$S_{1} = \left\{1\right\}, S_{2} = \left\{2, 3, 4\right\}, S_{3} = \left\{5, 6, 7, 8, 9\right\}, ...$, then the first element of $S_{21}$ is

Sequences and Series

Solution:

The given groups of numbers can be arranged as
image
Number of terms upto $S_{20} = 1 + 3 + 5 +.... $ to $20$ terms
$= \frac{20}{2} \left[2+\left(19\right)2\right] = 400$
$\therefore S_{12}$ starts with $401$.