Solution:
Given that,
Sn = 4n − n2
First term, a = S 1 = 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
Nth term, an = 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 the first two terms is 4. The second term is 1.
The 3rd, the 10th, and the nth terms are −1, −15, and 5 − 2n respectively.