You need to remember that you may evaluate the common difference if you know any two consecutive terms of arithmetic progression.

Notice that the problem provides the sum of n terms, hence you may find the sum of n-1 terms such that:

`S_(n-1) = 3(n-1)^2 - 2(n-1)`

You need to...

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You need to remember that you may evaluate the common difference if you know any two consecutive terms of arithmetic progression.

Notice that the problem provides the sum of n terms, hence you may find the sum of n-1 terms such that:

`S_(n-1) = 3(n-1)^2 - 2(n-1)`

You need to expand binomial such that:

`S_(n-1) = 3n^2 - 6n + 3 - 2n + 2`

Collecting like terms yields:

`S_(n-1) = 3n^2 - 8n + 5`

You need to remember that you may find the n-th term `a_n ` such that:

`S_n - S_(n-1) = 3n^2 - 2n - 3n^2+ 8n- 5 = a_n`

Reducing like terms yields:

`S_n - S_(n-1) = 6n-5`

Hence, you may evaluate `a_(n-1) ` such that:

`a_(n-1) = 6(n-1) - 5`

`a_(n-1) = 6n - 6 - 5`

`a_(n-1) = 6n - 11`

You need to subtract `a_(n-1) ` from `a_n` to find the common difference such that:

`a_n - a_(n-1) = 6n - 5 - 6n + 11`

Reducing like terms yields:

`a_n - a_(n-1) = 6`

**Hence, evaluating the common difference yields d = 6.**