The sum of the first \( n \) terms of an arithmetic sequence is given by \( S_n = 3n^2 + 5n \). What is the \( n \)-th term of the sequence?

["Understanding the Sum and the ( n )-th Term of an Arithmetic Sequence", "In mathematics, particularly in sequences and series, understanding key formulas helps unlock deeper insights into patterns and behaviors. One such important formula relates to the sum of the first ( n ) terms of an arithmetic sequence. In this article, we explore a specific known sum formula and derive the corresponding ( n )-th term of the sequence.", "---", "### Given:\nThe sum of the first ( n ) terms of an arithmetic sequence is:", "[\nS_n = 3n^2 + 5n\n]", "---", "### Objective:\nFind the ( n )-th term ( a_n ) of the arithmetic sequence using the given sum formula.", "---", "### Step 1: Recall the Standard Relation for Arithmetic Sequences", "For any arithmetic sequence, the sum of the first ( n ) terms can also be expressed in two equivalent ways:", "- Formula 1: ( S_n = \frac{n}{2} (2a + (n-1)d) )\n- Formula 2: ( S_n = \frac{n}{2} [2a_1 + (n-1)d] ), where ( a = a_1 ) is the first term and ( d ) is the common difference", "Our goal is to extract ( a_n = a + (n-1)d ), the ( n )-th term.", "---", "### Step 2: Use the Difference Between Consecutive Sums", "A powerful technique is to find the ( n )-th term by computing the difference between successive sums:", "[\na_n = S_n - S_{n-1}\n]", "This works because ( S_n ) is the sum up to ( n ), ( S_{n-1} ) up to ( n-1 ), so their difference reveals the ( n )-th term.", "Compute ( S_{n-1} ):", "[\nS_{n-1} = 3(n-1)^2 + 5(n-1)\n]", "Expand:", "[\n= 3(n^2 - 2n + 1) + 5n - 5\n= 3n^2 - 6n + 3 + 5n - 5\n= 3n^2 - n - 2\n]", "Now compute the difference:", "[\na_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2)\n]", "Simplify:", "[\n= 3n^2 + 5n - 3n^2 + n + 2\n= 6n + 2\n]", "---", "### Step 3: Identify the First Term and Common Difference", "We now have:", "[\na_n = 6n + 2\n]", "From the standard arithmetic sequence formula:", "[\na_n = a + (n-1)d\n]", "Expanding:", "[\na_n = dn + (a - d)\n]", "Comparing coefficients with ( a_n = 6n + 2 ):", "- ( d = 6 ) (common difference)\n- ( a + (1 - 1)d = a = 2 ) ⇒ first term ( a = 2 )", "So the ( n )-th term is:", "[\na_n = 6n + 2\n]", "Alternatively, written in standard arithmetic form:", "[\na_n = 6n + (2) = a + (n-1) \cdot 6 \quad \Rightarrow \quad a = 2\n]", "---", "### Step 4: Verification via Direct Sum Formula", "As a check, compute ( S_1 ) and ( S_2 ), and compare with direct sum from ( a_n = 6n + 2 ):", "- ( a_1 = 6(1) + 2 = 8 ), so ( S_1 = 8 )\n- ( a_2 = 6(2) + 2 = 14 ), so ( S_2 = a_1 + a_2 = 8 + 14 = 22 )", "Now use given formula:", "- ( S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8 ) ✅\n- ( S_2 = 3(4) + 5(2) = 12 + 10 = 22 ) ✅", "Now compute ( a_2 ) via ( S_2 - S_1 ):", "[\na_2 = 22 - 8 = 14 \quad \ ext{matches } a + 6 = 2 + 6 = 14\n]", "The result is consistent.", "---", "### Summary", "Given the sum formula:", "[\nS_n = 3n^2 + 5n\n]", "- The ( n )-th term of the arithmetic sequence is:", "[\n\boxed{a_n = 6n + 2}\n]", "This sequence has a constant common difference ( d = 6 ), first term ( a = 2 ), and the sum behavior perfectly matches the quadratic formula, illustrating the deep connection between sequence sums and individual terms.", "---", "### Why This Matters", "This example demonstrates how algebraic manipulation of known sum formulas allows calculation of otherwise inaccessible terms in arithmetic sequences. Understanding this relationship strengthens problem-solving skills in algebra, calculus, and beyond—especially when dealing with discrete data or series expansions.", "If you're working with sequences, mastering such techniques turns abstract summations into tangible formulas. Keep practicing with sums and term differences—your intuition for sequences will grow significantly!", "---", "Keywords: arithmetic sequence sum formula, derive ( n )-th term, ( S_n = 3n^2 + 5n ), arithmetic progression, sum to terms conversion, ( a_n = S_n - S_{n-1} )"]