Given S_n = 4n – n²,

On putting n = 1, we get the first term(a), S₁ = a = 4(1) – 1² = 3.

On putting n = 2 gives S₂ = a + a + d = 4(2) – 2² = 4

=> d = 4 – 2a

=> d = 4 – 6 = –2

Second term(a₂) = a + d = 3 + (–2) = 1

By using the formula of nth term of an A.P.

a_n = a + (n – 1)d

So,

= 3 + (n – 1)(–2)

= 5 – 2n

Hence, third term(a₃) = 5 – 2(3) = –1

And tenth term(a₁₀) = 5 – 2(10) = –15

Hence, the first term is 3, sum of first two terms is 4, second term is 1 

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