Pregunta
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11 The sum to \( n \) terms of an arithmetic sequence is given by: \( S_{n}=63 n-3 n^{2} \) a Calculate the 4th term. b Determine the value(s) of \( n \) if the sum of \( n \) terms is -300 .

Ask by Munoz French. in South Africa
Jan 23,2025

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**Part a:** The 4th term is **42**. **Part b:** The value of \( n \) is **25**.

Solución

Let's solve the problem step-by-step. **Given:** The sum of the first \( n \) terms of an arithmetic sequence is: \[ S_{n} = 63n - 3n^{2} \] ### Part a: Calculate the 4th Term (\( t_4 \)) **Step 1: Understand the Relationship** For an arithmetic sequence: \[ S_{n} = \frac{n}{2} [2a + (n-1)d] \] where: - \( a \) is the first term. - \( d \) is the common difference. - \( t_n = a + (n-1)d \) is the \( n^{th} \) term. **Step 2: Equate the Given Sum to the Arithmetic Sum Formula** \[ \frac{n}{2} [2a + (n-1)d] = 63n - 3n^{2} \] **Step 3: Simplify the Equation** Divide both sides by \( n \) (assuming \( n \neq 0 \)): \[ \frac{1}{2} [2a + (n-1)d] = 63 - 3n \] Multiply both sides by 2: \[ 2a + (n - 1)d = 126 - 6n \] \[ 2a + dn - d = 126 - 6n \] **Step 4: Equate Coefficients** For the equation to hold for all \( n \), the coefficients of like terms must be equal: - Coefficient of \( n \): \[ d = -6 \] - Constant term: \[ 2a - d = 126 \] Substitute \( d = -6 \): \[ 2a - (-6) = 126 \\ 2a + 6 = 126 \\ 2a = 120 \\ a = 60 \] **Step 5: Find the 4th Term** \[ t_4 = a + (4-1)d = 60 + 3(-6) = 60 - 18 = 42 \] **Answer to Part a:** The 4th term is **42**. --- ### Part b: Determine the Value(s) of \( n \) if \( S_n = -300 \) **Step 1: Set Up the Equation** \[ 63n - 3n^{2} = -300 \] \[ -3n^{2} + 63n + 300 = 0 \] **Step 2: Simplify the Equation** Divide the entire equation by -3 to simplify: \[ n^{2} - 21n - 100 = 0 \] **Step 3: Solve the Quadratic Equation** Use the quadratic formula: \[ n = \frac{21 \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot (-100)}}{2 \cdot 1} \\ n = \frac{21 \pm \sqrt{441 + 400}}{2} \\ n = \frac{21 \pm \sqrt{841}}{2} \\ n = \frac{21 \pm 29}{2} \] This gives two solutions: 1. \( n = \frac{21 + 29}{2} = \frac{50}{2} = 25 \) 2. \( n = \frac{21 - 29}{2} = \frac{-8}{2} = -4 \) **Step 4: Determine Validity** Since \( n \) represents the number of terms, it must be a positive integer. **Answer to Part b:** The valid value of \( n \) is **25**.

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Bonus Knowledge

To calculate the 4th term of the arithmetic sequence, we can use the formula for the sum of \( n \) terms, \( S_n = 63n - 3n^2 \). The \( n \)th term, \( a_n \), can be found using the relation \( a_n = S_n - S_{n-1} \). First, calculate \( S_4 \): \[ S_4 = 63(4) - 3(4^2) = 252 - 48 = 204. \] Next, calculate \( S_3 \): \[ S_3 = 63(3) - 3(3^2) = 189 - 27 = 162. \] Now, find the 4th term: \[ a_4 = S_4 - S_3 = 204 - 162 = 42. \] So the 4th term is \( 42 \). Next, to find \( n \) when \( S_n = -300 \): Set up the equation: \[ 63n - 3n^2 = -300. \] Rearranging gives: \[ 3n^2 - 63n - 300 = 0. \] Dividing by 3: \[ n^2 - 21n - 100 = 0. \] Now, apply the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ n = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(-100)}}{2(1)} = \frac{21 \pm \sqrt{441 + 400}}{2} = \frac{21 \pm \sqrt{841}}{2} = \frac{21 \pm 29}{2}. \] This results in: \[ n = \frac{50}{2} = 25 \quad \text{and} \quad n = \frac{-8}{2} = -4. \] Since \( n \) must be a positive integer, the valid value is \( n = 25 \). Summarizing: a) The 4th term is \( 42 \). b) The value of \( n \) for which the sum of \( n \) terms is -300 is \( n = 25 \).

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