Step-by-step explanation:

Given, sum of first n terms of an AP, Sn = 3n2 – n

Sn = 3n2 – n

Replacing n by n – 1, we get

Let the nth term of AP be an.

Now,put n = 1 in an, we get a = 6×1−4 = 2put n = 2 in an, we get a2 = 6×2−4 = 8put n = 3 in an, we get a3 = 6×3−4 = 14and so on.So, required AP is, 2,8,14,.....

Putting n = 50 we get

a50 = 6 × 50– 4 = 300– 4 = 296

Thus, the 50th term of the given A.P. is 296

Answer:

9

Step-by-step explanation:

an= Sn-Sn-1

50 = 3n^2-n-(3(n-1)^2-(n-1))

50=3n^2-n-(3(n^2-2n+1)-n+1)

50=3n^2-n-3n^2+6n-3+n-1

50=6n-4

6n=54

n=54/6

n=9