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The common difference is a critical concept in arithmetic sequences. It's the fixed amount that each term increases (or decreases) by as you move from one term to the next. In our problem, the common difference (\(d\)) is 3. This means every consecutive term in the sequence is 3 units larger than the one before it.
Understanding the common difference helps in predicting future terms in the sequence without listing them all. For example, given the first term (\(a_1 = 6\)) and the common difference (\(d = 3\)), the second term is simply \(a_1 + d = 6 + 3 = 9\), the third term is \(9 + 3 = 12\), and so forth. This steady increment forms the basis for calculating any term in the sequence.

The formula to find the n-th term of an arithmetic sequence is \(a_n = a_1 + (n - 1)d\). This formula is very handy because it allows you to directly calculate any term in the sequence without having to list all the preceding terms.
In the given problem, we needed to find the 6th term (\(a_6\)). Plugging in the values, we use \(a_6 = 6 + (6 - 1)3 = 6 + 15 = 21\). This means the 6th term in the sequence is 21. This formula shows how the first term (\(a_1\)) and the common difference (\(d\)) play a crucial role in determining any term in the sequence.

To find the sum of the first n terms (\(S_n\)) of an arithmetic sequence, use the formula \(S_n = \frac{n}{2} (a_1 + a_n)\). This formula essentially adds the first term to the n-th term and then multiplies by half the number of terms. Think of it as averaging the first and last term and then multiplying by the number of terms.
In this exercise, we wanted to find the sum of the first 6 terms. First, we calculated the 6th term (\(a_6 = 21\)). Plugging the values into the sum formula gives us \(S_6 = \frac{6}{2} (6 + 21) = 3 \times 27 = 81\). Hence, the sum of the first 6 terms in this series is 81. This formula helps in quickly summing up a large number of terms without adding each one individually.