The sum of the first n terms of an arithmetic sequence is \( S_n = 120 \), with \( a = 4 \), \( d = 4 \). Find n.

["Title: How to Find the Number of Terms ( n ) in an Arithmetic Sequence Given the Sum", "Meta Description:\nLearn how to calculate the number of terms ( n ) in an arithmetic sequence when the sum of the first ( n ) terms is known. Given ( S_n = 120 ), ( a = 4 ), and ( d = 4 ), solve for ( n ) using the arithmetic sum formula.", "---", "## Introduction", "Understanding how to find the number of terms in an arithmetic sequence is essential for solving problems in mathematics, engineering, and data analysis. One common question is: Given the sum ( S_n = 120 ), initial term ( a = 4 ), and common difference ( d = 4 ), find the number of terms ( n )? In this article, we’ll walk through the formula and solve step-by-step.", "---", "## What is the Sum of the First ( n ) Terms?", "The sum of the first ( n ) terms of an arithmetic sequence is given by:", "[\nS_n = \frac{n}{2} \left( 2a + (n - 1)d \right)\n]", "Where:\n- ( S_n ) = sum of first ( n ) terms\n- ( a ) = first term\n- ( d ) = common difference\n- ( n ) = number of terms (what we are solving for)", "---", "## Given Values", "We are told:\n[\nS_n = 120,\quad a = 4,\quad d = 4\n]", "Plug these into the sum formula:", "[\n120 = \frac{n}{2} \left( 2 \cdot 4 + (n - 1) \cdot 4 \right)\n]", "---", "## Step-by-Step Solution", "### 1. Simplify inside the parentheses", "Calculate constants:", "[\n2 \cdot 4 = 8,\quad (n - 1) \cdot 4 = 4(n - 1)\n]", "So:", "[\nS_n = \frac{n}{2} \left( 8 + 4(n - 1) \right)\n]", "### 2. Expand the expression inside the parentheses", "[\n8 + 4(n - 1) = 8 + 4n - 4 = 4n + 4\n]", "Now the equation becomes:", "[\n120 = \frac{n}{2} (4n + 4)\n]", "### 3. Multiply both sides by 2 to eliminate the fraction", "[\n240 = n (4n + 4)\n]", "### 4. Expand the right-hand side", "[\n240 = 4n^2 + 4n\n]", "### 5. Rearrange into standard quadratic form", "[\n4n^2 + 4n - 240 = 0\n]", "Divide the entire equation by 4 to simplify:", "[\nn^2 + n - 60 = 0\n]", "---", "## Solve the Quadratic Equation", "Use the quadratic formula:", "[\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For ( n^2 + n - 60 = 0 ), coefficients are:\n- ( a = 1 )\n- ( b = 1 )\n- ( c = -60 )", "Compute discriminant:", "[\n\Delta = 1^2 - 4(1)(-60) = 1 + 240 = 241\n]", "Now compute:", "[\nn = \frac{-1 \pm \sqrt{241}}{2}\n]", "Since ( \sqrt{241} \approx 15.52 ), we get:", "[\nn = \frac{-1 + 15.52}{2} \approx \frac{14.52}{2} \approx 7.26\n]\n[\nn = \frac{-1 - 15.52}{2} \approx \frac{-16.52}{2} = -8.26 \quad (\ ext{discard, since } n > 0)\n]", "But we expect an integer solution since ( n ) must be a positive whole number. Let’s check if 8 works.", "---", "## Testing Integer Values Close to Solution", "Try ( n = 8 ):", "[\nS_8 = \frac{8}{2} (2 \cdot 4 + (8 - 1) \cdot 4) = 4 (8 + 28) = 4 \cdot 36 = 144 \quad (\ ext{Too high})\n]", "Try ( n = 6 ):", "[\nS_6 = \frac{6}{2} (8 + 5 \cdot 4) = 3 (8 + 20) = 3 \cdot 28 = 84 \quad (\ ext{Too low})\n]", "Try ( n = 7 ):", "[\nS_7 = \frac{7}{2} (8 + 6 \cdot 4) = \frac{7}{2} (8 + 24) = \frac{7}{2} \cdot 32 = 7 \cdot 16 = 112 \quad (\ ext{Still low})\n]", "Try ( n = 8 ): already 144 (too high)", "But wait — earlier quadratic gave approximate 7.26. Maybe we made an early assumption error?", "Wait — go back to:", "[\n240 = n(4n + 4) = 4n^2 + 4n\n]", "Try factoring original quadratic accurately:", "[\n4n^2 + 4n - 240 = 0 \quad \Rightarrow \quad n^2 + n - 60 = 0\n]", "Check perfect squares:", "Is ( n^2 + n - 60 ) factorable?", "Find two numbers that multiply to ( -60 ) and add to ( 1 ):\n( 8 \ imes (-7) = -56 ) → no\n( 8 \ imes (-6) = -48 ) → no\n( 10 \ imes (-6) = -60 ), ( 10 - 6 = 4 ) → no\n( 15 \ imes (-4) = -60 ), ( 15 - 4 = 11 ) → no\n( 8 \ imes (-7.5)? ) not integer.", "But approximation was ( n \approx 7.26 ), not integer — contradiction?", "Wait — double-check sum calculation:", "Let’s compute sum at ( n = 6, 7, 8 ) directly:", "- ( a = 4, d = 4 ) → sequence: 4, 8, 12, 16, 20, 24, ...", "Try ( n = 6 ):\nSum = ( \frac{6}{2}(2\cdot4 + 5\cdot4) = 3(8 + 20) = 3(28) = 84 )", "( n = 7 ):\n( S_7 = \frac{7}{2}(8 + 6\cdot4) = \frac{7}{2}(32) = 112 )", "( n = 8 ):\n( S_8 = \frac{8}{2}(8 + 7\cdot4) = 4(8 + 28) = 4 \cdot 36 = 144 )", "But given ( S_n = 120 ), which is between 112 (n=7) and 144 (n=8). Not achievable with integer ( n )? Contradiction?", "Wait — but our quadratic gave non-integer root. That suggests no integer ( n ) satisfies the condition exactly.", "But problem states sum is 120 — let’s solve the quadratic exactly.", "Back to:", "[\nn^2 + n - 60 = 0\n]", "[\nn = \frac{-1 \pm \sqrt{241}}{2}\n]", "Since ( \sqrt{241} ) is irrational, no integer ( n ) satisfies ( S_n = 120 ) exactly with ( a = 4 ), ( d = 4 ).", "But perhaps there's a typo? Or is this a trick question?", "Wait — double-check discriminant step:", "We had:", "[\n240 = 4n^2 + 4n \Rightarrow 4n^2 + 4n - 240 = 0 \Rightarrow n^2 + n - 60 = 0\n]", "Correct.", "Now compute discriminant:", "[\n\Delta = 1 + 240 = 241 \quad (\ ext{not a perfect square})\n]", "So no integer solution exists.", "But maybe the problem assumes rounding? Or miscalculation?", "Wait — recompute ( S_7 ):", "Terms: 4, 8, 12, 16, 20, 24, 28 → 7 terms", "Sum = ( \frac{7}{2}(4 + 28) = \frac{7}{2} \cdot 32 = 112 )", "( S_8 = 4 + 8 + 12 + 16 + 20 + 24 + 28 + 32 = )", "Add step-by-step:\n4 + 8 = 12\n12 + 12 = 24\n24 + 16 = 40\n40 + 20 = 60\n60 + 24 = 84\n84 + 28 = 112\n112 + 32 = 144", "Yes, 144.", "So sum increases from 112 at ( n=7 ) to 144 at ( n=8 ), skipping 120. Therefore, no integer ( n ) satisfies ( S_n = 120 ).", "But if the problem states ( S_n = 120 ), and ( a=4, d=4 ), then no solution in positive integers.", "However, suppose the problem allows approximate ( n ), then:", "From quadratic:", "[\nn = \frac{-1 + \sqrt{241}}{2} \approx \frac{-1 + 15.524}{2} \approx 7.262\n]", "But since ( n ) must be integer, and sum is discrete, no such ( n ) exists.", "---", "## Correction: ReChecking the Sum Formula", "Wait — is the formula correctly applied?", "Given:\n( a = 4 ), ( d = 4 ) → sequence: 4, 8, 12, 16,...", "Sum formula:\n[\nS_n = \frac{n}{2} [2a + (n - 1)d] = \frac{n}{2} [8 + 4(n - 1)] = \frac{n}{2} (4n + 4) = 2n(n + 1)\n]", "Yes — simplifies to ( S_n = 2n(n + 1) )", "Set equal to 120:", "["]