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

["Title: Deriving the Common Difference of an Arithmetic Sequence from Its Sum Formula", "Understanding the sum of an arithmetic sequence is fundamental in algebra, and recognizing how to extract key properties—such as the common difference—from the sum formula unlocks deeper insights into sequence behavior. In this article, we explore how the given sum formula ( S_n = 3n^2 + 5n ) reveals the common difference of an arithmetic sequence.", "### The Sum of the First ( n ) Terms of an Arithmetic Sequence", "For any arithmetic sequence where the first term is ( a ) and the common difference is ( d ), the sum of the first ( n ) terms is typically expressed as:\n[\nS_n = \frac{n}{2} \left(2a + (n-1)d\right)\n]\nThis formula derives from averaging the first and ( n )-th terms: ( S_n = \frac{n}{2}(a + l) ), with ( l = a + (n-1)d ).", "Interestingly, the problem gives the sum in a quadratic form:\n[\nS_n = 3n^2 + 5n\n]\nThis quadratic expression implies the sequence is quadratic—typically characteristic of arithmetic sequences only when transformed via partial summation.", "### Uncovering the Sequence via Difference of Sums", "To find the common difference ( d ), we use the key idea: the sum of terms from ( k = 1 ) to ( n ) equals ( S_n ), so the ( n )-th partial sum is:\n[\ns_n = S_n - S_{n-1}\n]", "Compute ( s_n ) from the given formula:\n[\ns_n = (3n^2 + 5n) - \left[3(n-1)^2 + 5(n-1)\right]\n]\nExpand ( S_{n-1} ):\n[\nS_{n-1} = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2\n]\nNow compute the difference:\n[\ns_n = (3n^2 + 5n) - (3n^2 - n - 2) = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2\n]", "Thus, the sum of the first ( n ) terms is:\n[\ns_n = 6n + 2\n]", "### Extracting the Arithmetic Sequence", "Since ( s_n = \frac{n}{2}(2a + (n-1)d) ), let’s expand it:\n[\ns_n = \frac{n}{2}(2a - d + nd) = \frac{d}{2}n^2 + \left(a - \frac{d}{2}\right)n\n]\nCompare this with the derived sum ( s_n = 6n + 2 ), which has no ( n^2 ) term:\n[\n\frac{d}{2} = 0 \quad \Rightarrow \quad d = 0\n]\nWait—this suggests ( d = 0 ), but that would make the sequence constant, which contradicts the quadratic growth unless verified carefully.", "But hold: ( s_n = 6n + 2 ) is linear in ( n ), yet the general formula for arithmetic sequence sum has degree 2 if the sequence is truly arithmetic (with ( d <br/>\ne 0 )). However, a linear sum corresponds to a constant sequence (when ( d = 0 )), unless the coefficient of ( n^2 ) vanishes.", "Here, ( s_n = 6n + 2 ) has no ( n^2 ) term, so ( \frac{d}{2} = 0 \Rightarrow d = 0 ). But a constant sequence satisfies ( S_n = na ), which is linear.", "Is this sequence constant? Let’s test:\n- ( s_1 = 6(1) + 2 = 8 ) → ( a = 8 )\n- ( s_2 = 6(2) + 2 = 14 ) → sum of first two terms: ( 8 + (8 + d) = 16 + d )\nBut ( s_2 = 14 ), so ( 16 + d = 14 \Rightarrow d = -2 )", "Contradiction? Wait — we derived ( s_n = 6n + 2 ), so:\n- ( s_1 = 8 )\n- ( s_2 = 14 )\n- ( s_3 = 3(9) + 5(3) = 27 + 15 = 42 )", "Now compute differences:\n- ( s_1 = 8 )\n- ( s_2 - s_1 = 14 - 8 = 6 )\n- ( s_3 - s_2 = 42 - 14 = 28 )", "But for arithmetic sequence, differences of partial sums should be constant. Here, differences are ( 6 ), ( 28 ), not equal—unless our earlier assumption needs correction.", "Wait — reconsider: earlier calculation gave ( s_n = 6n + 2 ), but let's recheck algebra:", "Original:\n[\nS_n = 3n^2 + 5n\n]\n[\nS_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2\n]\nThen:\n[\ns_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2) = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2\n]\nSo correct: ( s_n = 6n + 2 )", "Now, in an arithmetic sequence, the ( n )-th term ( a_n = s_n - s_{n-1} ) for ( n \ge 2 ). But here:\n[\na_n = s_n - s_{n-1} = (6n + 2) - [6(n-1) + 2] = 6n + 2 - (6n - 6 + 2) = 6n + 2 - 6n + 4 = 6\n]\nSo for ( n \ge 2 ), ( a_n = 6 ) — constant.", "Then ( a_1 = s_1 = 6(1) + 2 = 8 )", "Thus, the sequence is:\n[\n8, 6, 6, 6, 6, \dots\n]\nThis is not arithmetic unless interrupted—but a constant sequence has common difference ( d = 0 ). But the first term is 8, all others 6? No—( a_2 = s_2 - s_1 = 14 - 8 = 6 ), ( a_3 = 42 - 14 = 28 )? Contradiction.", "Wait: ( s_2 = 3(4) + 5(2) = 12 + 10 = 22 ), not 14!", "Ah! Critical error: the formula ( S_n = 3n^2 + 5n ) evaluated at ( n = 2 ):\n[\nS_2 = 3(4) + 5(2) = 12 + 10 = 22\n]\nBut earlier we claimed ( s_2 = 14 )—this is false.", "Correct evaluation:\n[\nS_n = 3n^2 + 5n\n]\nSo:\n- ( S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8 ) → ( a_1 = 8 )\n- ( S_2 = 3(4) + 10 = 12 + 10 = 22 ) → ( a_2 = S_2 - S_1 = 22 - 8 = 14 )\n- ( S_3 = 3(9) + 15 = 27 + 15 = 42 ) → ( a_3 = 42 - 22 = 20 )\n- ( S_4 = 3(16) + 20 = 48 + 20 = 68 ) → ( a_4 = 68 - 42 = 26 )", "Sequence: ( 8, 14, 20, 26, \dots ) — clearly increasing by 6 each time.", "So ( a_n = 6n + 2 )? Check:\n- ( n = 1 ): ( 6(1) + 2 = 8 ) ✓\n- ( n = 2 ): ( 12 + 2 = 14 ) ✓\n- ( n = 3 ): ( 18 + 2 = 20 ) ✓\n- ( n = 4 ): ( 24 + 2 = 26 ) ✓", "Yes! ( a_n = 6n + 2 )", "But for an arithmetic sequence, ( a_n = a + (n-1)d ), linear in ( n ), which matches.", "Now, standard sum:\n[\nS_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (6k + 2) = 6\sum k + 2\sum 1 = 6\cdot\frac{n(n+1)}{2} + 2n = 3n(n+1) + 2n = 3n^2 + 3n + 2n = 3n^2 + 5n\n]\nExactly matches the given sum!", "Thus, the sequence is arithmetic with first term ( a = 8 ), common difference ( d = a_2 - a_1 = 14 - 8 = 6 ).", "Verify:\n[\na_n = a + (n-1)d = 8 + (n-1)(6) = 8 + 6n - 6 = 6n + 2\n]\nPerfect.", "### Conclusion: The Common Difference", "From the quadratic form ( S_n = 3n^2 + 5n ), partial summation yields a linear sequence ( s_n = 6n + 2 ), implying the arithmetic sequence increases by a constant difference. Computing the difference between consecutive terms confirms:\n[\nd = a_2 - a_1 = 14 - 8 = 6\n]\nThus, the common difference of the sequence is ( \boxed{6} ).", "### Why This Matters", "Recognizing sum formulas allows reverse-engineering of sequence properties. The absence of a quadratic coefficient in the sum implies linearity, but only when the expression simplifies appropriately. Here, diff attentiveness reveals the true structure—proving ( d = 6 ), not zero.", "This method extends to any sum formula: identify the polynomial degree, use difference to extract leading coefficients, and derive ( d ) from the linear term in the arithmetic sum.", "---\nKeywords: arithmetic sequence sum formula, ( S_n = 3n^2 + 5n ), common difference, derive ( d ) from sum, partial summation, sequence analysis", "Meta Description: Learn how the quadratic sum ( S_n = 3n^2 + 5n ) reveals the common difference ( d = 6 ) of an arithmetic sequence through partial summation and term comparison."]