If the sum of the first n terms of an AP is \(4n - n^2\), what is the first term (that is \(S_1\))? What is the sum of first two terms? What is the second term? Similarly find the \(3^{rd}\), the \(10^{th}\) and the nth terms.

Given that,
\(S_n = 4n-n^2\)
First term, a = \(S_1 = 4(1) - (1)2 = 4-1 = 3\)
Sum of first two terms = \(S_2= 4(2)-(2)2 = 8-4 = 4\)
Second term, \(a_2 = S_2 - S_1 = 4-3 = 1\)
Common difference, d = \(a_2-a = 1-3 = -2\)

Nth term, \(a_n = a+(n-1)d \)
= 3 + (n -1)(-2)
= 3 - 2n + 2
= 5 - 2n

Therefore,\( a_3 = 5 - 2(3) = 5 - 6 = -1\)
\(a_{10} = 5-2(10) = 5 - 20 = -15\)

Hence, the sum of first two terms is 4. The second term is 1.

The 3rd, the 10th, and the nth terms are -1, -15, and 5 - 2n respectively.