A new sequence is obtained from the sequence of positive integers (1, 2, 3,...) by deleting all the perfect squares. Then the $$2022^{nd}$$ term of the new sequence is___________.

We know. $$45^2=2025$$

Now, if the series was 1,2,3,4,.....,2026,..... then 2026 number would have been $$2026^{th}$$ term only

But since we are eliminating all perfect squares, so the number 2026 has shifted 45 places forward.

(It shifts 45 places forward because before 2026, there are 45 perfect squares, as $$45^2=2025$$)

So, now, 2026 is the $$(2026-45)^{th}$$ term = $$1981^{st}$$ term

So, $$2022^{nd}$$ term =$$(1981+41)^{th}$$term = 41 numbers after $$1981^{th}$$ term = 2026+41=2067