Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 3n² + 5n

Then n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (3n² + 5n) - [3(n - 1)² + 5(n - 1)]

= (3n² + 5n) - [3(n² + 1 - 2n) + 5n - 5]

= 3n² + 5n - 3n² - 3 + 6n - 5n + 5

= 2 + 6n

Now, common difference = d = a_n - a_{n - 1}

= 2 + 6n - [2 + 6(n - 1)]

= 2 + 6n - 2 - 6n + 6

= 6

∴ Common difference = 6

ALITER: Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 3n² + 5n

Put n = 1, we get S₁ = 8

Put n = 2, we get S₂ = 22

Now S₁ = a₁

a₂ = S₂ - S₁

∴ a₂ = 22 - 8 = 14

Now, d = a₂ - a₁

= 14 - 8

= 6

∴ Common difference = 6