In an AP :

given a = 5, d = 3, a_n = 50 find n and s_n

d is the common difference = 3

a is the first term = 5

a_n = 50

a_n = a + (n-1)d

50 = 5 + (n-1)(3)

45 = 3(n-1)

15 = n-1

n = 16

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

= [ 2×5 + (16-1)(3)]

= 8 [ 55 ]

⇒ 440