In an AP :

given a = 2, S_n = 90, d = 8 find n and a_n

First term (a) = 2

Common difference (d) = 8

S_n = 90

The sum of the series is S_n = [2a + (n-1) d]

90 = [2×2 + (n-1)8]

90 = 4n² – 2n

2n² – n - 45 = 0

By solving the above quadratic equation we will get

2n² -10n + 9n – 45 = 0

2n(n-5) + 9(n-5) = 0

n = 5

a_n = a + (n-1)d

a₅ = 2 + (5-1)8 = 34