Given the arithmetic sequence 6, 10, 14, 18, ..., find the formula for the n-th term and the sum of the first n terms.

• The sequence is $6, 10, 14, 18, \dots$
• First term, $a = 6$.
• Common difference, $d = 10 - 6 = 4$.

• The formula for the $n$-th term of an arithmetic sequence is $U_n = a + (n-1)d$.
• Substituting the values: $$U_n = 6 + (n-1)4$$ $$U_n = 6 + 4n - 4$$ $$U_n = 4n + 2$$

This is correct.

• The formula for the sum of the first $n$ terms of an arithmetic sequence is $S_n = \frac{n}{2}(2a + (n-1)d)$.
• Substituting the values: $$S_n = \frac{n}{2}(2(6) + (n-1)4)$$ $$S_n = \frac{n}{2}(12 + 4n - 4)$$ $$S_n = \frac{n}{2}(4n + 8)$$ $$S_n = n(2n + 4)$$ $$S_n = 2n^2 + 4n$$

This is also correct.

Your solution matches the correct derivation for both the $n$-th term and the sum of the first $n$ terms of the given arithmetic sequence.