got confused at a point while solving this AP. somebody pls help?

Question

Find the least number of terms of the AP, 1+3+5+.... that are required to make a sum exceeding 4000

Answer 1

This is the AP of odd numbers. The ninth term is a(n) = 2n -1, and, for each n, S(n) = 1 + 3 + .....2n -1 = ((a_1 + a_n)*n)/2 = 2n^2/2 = n^2. Therefore, you have to find the least positive integer number n such that n^2 > 4000, which means the least positive n > sqrt(4000) = 10 sqrt(40) Since 63 < sqrt(4000) < 64, we have n = 64 .

Answer 2

in formula Sn=n/2(2a+(n-1)d) we get, if Sn=4000
4000=n^2
it is clear that we have to search for the least number having its square greater than 4000, hence the answer is 64 because 64x64=4096

Answer 3

n/2 x [2(1)+ (n-1)x2] > 4000
n x (2 + 2n - 2) > 8000
2n^2>8000
n>63.2 or n<0 (rej)

least n is 64