Given that,
\(S_n = 4n − n^2\)
First term, \(a_1 = S_1 = 4(1) − (1)^2 = 4 − 1 = 3\)
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
\(d = a_2 − a_1 = 1 − 3 = −2\)
\(a_n = a + (n − 1)d \)
\(a_n = 3 + (n − 1) (−2)\)
\(a_n = 3 − 2n + 2\)
\(a_n = 5 − 2n\)
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. 3rd, 10th, and nth terms are −1, −15, and 5−2n respectively.