Given that,

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.