Given, S_n = 4n−n²

First term, a = S₁ = 4(1) − (1)² 

= 4−1 = 3

Sum of first two terms = S₂= 4(2)−(2)² 

= 8−4 = 4

Second term, a₂ = S₂ − S₁ 

= 4−3 = 1

Common difference, d = a₂−a 

= 1−3 = −2

N^th term, a_n = a+(n−1)d 

= 3+(n −1)(−2)

= 3−2n +2

= 5−2n

Therefore, a₃ = 5−2(3)

= 5-6 = −1

a₁₀ = 5−2(10)

= 5−20 = -15

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

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