Solution:
Provided here that,
Sn = 4n−n2
The first term, a = S1 = 4(1) − (1)2 = 4−1 = 3
The sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4
The second term, a2 = S2 − S1 = 4−3 = 1
The common difference, d = a2−a = 1−3 = −2
Nth term, an = a+(n−1)d
= 3+(n −1)(−2)
= 3−2n +2
= 5−2n
As a result, a3 = 5−2(3) = 5-6 = −1
a10 = 5−2(10) = 5−20 = −15
Hence, as a result the sum of first two terms is 4. The second term is 1.
−1, −15, and 5 − 2n are the values of the 3rd, the 10th, and the nth terms, respectively.