Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = [2a + (n − 1)d]
It is given that sum of the first n terms of an A.P. is 5n − n2.
∴ First term = a = S1 = 5(1) − (1)2 = 4.
Sum of first two terms = S2 = 5(2) − (2)2 = 6.
∴ Second term = S2 − S1 = 6 − 4 = 2.
∴ Common difference = d = Second term − First term
= 2 − 4 = −2
Also, nth term = an = a + (n − 1)d
⇒ an = 4 + (n − 1)(−2)
⇒ an = 4 − 2n + 2
⇒ an = 6 − 2n
Thus, nth term of this A.P. is 6 − 2n.