Find the 51st term of the series.

A particular series is given by :

A1 = {1}
A2 = {2,3,4}
A3 = {5,6,7,8,9}
A4 = {10,11,12,13,14,15,16}
and so on...

Find the first term of the A51.



I'll give it a try.

Here each term has basically a set of integers.
Careful observation reveals 2 things:

1: Each term has last element of it its set as the square of the term number. That is for A2, say n =2 then last element of A2 is n^2=4. Similarly for A3, last element of set is 3^3=9.
2: The number of elements in the set of nth term = n+(n-1). For example for A2, n=2 and number of elements in A2 = 2+1=3. Similarly for A3, n=3 hence number of elements in set of A3 = 3+2 =5.

These 2 observations reveal that A51 will have 51+50 = 101 terms and the last term of A51 will be 51^2 and moving backwards each previous terms would be 1 less.

Hence A51 = {........,.,.,.,,.,.....,(51^2-2), (51^2-1), (51^2) }

And the first term of A51 will be 51^2-100.

Please give kudos if this helps !