**Solution:**

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

N^{th} 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 the 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.