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The sum to \( n \) terms of an arithmetic series is \( S_{n}=\frac{\pi}{2}(7 n+15 \) 3.1 How many terms must be added to give a sum of 425 ? 3.2 Determine the sixth term of the series.

Ask by Guerrero Hall. in South Africa
Feb 26,2025

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3.1 Approximately 36 terms are needed to sum to 425. 3.2 The sixth term of the series is \( \frac{7\pi}{2} \).

Solución

To solve the problem, we start with the given sum of the first \( n \) terms of an arithmetic series: \[ S_n = \frac{\pi}{2}(7n + 15) \] ### Step 1: Find \( n \) for \( S_n = 425 \) We need to find \( n \) such that: \[ \frac{\pi}{2}(7n + 15) = 425 \] To isolate \( n \), we first multiply both sides by \( \frac{2}{\pi} \): \[ 7n + 15 = \frac{850}{\pi} \] Next, we subtract 15 from both sides: \[ 7n = \frac{850}{\pi} - 15 \] Now, we divide by 7 to solve for \( n \): \[ n = \frac{1}{7} \left( \frac{850}{\pi} - 15 \right) \] ### Step 2: Calculate \( n \) Now, we will calculate the value of \( n \) using the approximate value of \( \pi \approx 3.14 \): \[ n = \frac{1}{7} \left( \frac{850}{3.14} - 15 \right) \] Calculating \( \frac{850}{3.14} \): \[ \frac{850}{3.14} \approx 270.06 \] Now substituting back: \[ n = \frac{1}{7} (270.06 - 15) = \frac{1}{7} (255.06) \approx 36.43 \] Since \( n \) must be a whole number, we round \( n \) to the nearest whole number, which is 36. ### Step 3: Find the sixth term of the series The \( n \)-th term of an arithmetic series can be found using the formula: \[ a_n = S_n - S_{n-1} \] Where \( S_n \) is the sum of the first \( n \) terms and \( S_{n-1} \) is the sum of the first \( n-1 \) terms. First, we calculate \( S_6 \) and \( S_5 \): \[ S_6 = \frac{\pi}{2}(7 \cdot 6 + 15) = \frac{\pi}{2}(42 + 15) = \frac{\pi}{2}(57) = \frac{57\pi}{2} \] \[ S_5 = \frac{\pi}{2}(7 \cdot 5 + 15) = \frac{\pi}{2}(35 + 15) = \frac{\pi}{2}(50) = 25\pi \] Now, we find the sixth term: \[ a_6 = S_6 - S_5 = \frac{57\pi}{2} - 25\pi \] To combine these, we convert \( 25\pi \) to have a common denominator: \[ 25\pi = \frac{50\pi}{2} \] Now substituting back: \[ a_6 = \frac{57\pi}{2} - \frac{50\pi}{2} = \frac{7\pi}{2} \] ### Final Answers 3.1 The number of terms that must be added to give a sum of 425 is approximately \( n = 36 \). 3.2 The sixth term of the series is \( a_6 = \frac{7\pi}{2} \).

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To find how many terms must be added to give a sum of 425, we start by setting the equation for the sum \( S_n \) equal to 425: \[ S_n = \frac{\pi}{2}(7n + 15) = 425. \] Solving for \( n \): \[ 7n + 15 = \frac{425 \times 2}{\pi} \approx 270.17 \quad (\text{using } \pi \approx 3.14), \] \[ 7n = 270.17 - 15 = 255.17, \] \[ n \approx \frac{255.17}{7} \approx 36.5. \] Since \( n \) must be a whole number, we round up to 37. So, 37 terms must be added to achieve the sum of 425. For the sixth term of the series, we know the formula for the \( n \)-th term of an arithmetic series is: \[ a_n = a + (n-1)d, \] where \( a \) is the first term and \( d \) is the common difference. We can express \( a \) and \( d \) based on the sum \( S_n \): The first term can be derived from \( S_1 \): \[ S_1 = \frac{\pi}{2}(7 \cdot 1 + 15) = \frac{\pi}{2}(22) = 11\pi. \] Thus, the first term \( a \) is \( 11\pi \). To find \( d \), we can calculate \( S_2 \) for insight into the difference between consecutive terms: \[ S_2 = \frac{\pi}{2}(7 \cdot 2 + 15) = \frac{\pi}{2}(29) = \frac{29\pi}{2}. \] Calculating the second term: \[ a_2 = S_2 - S_1 = \frac{29\pi}{2} - 11\pi = \frac{29\pi}{2} - \frac{22\pi}{2} = \frac{7\pi}{2}. \] So, \( d = a_2 - a = \frac{7\pi}{2} - 11\pi = \frac{7\pi - 22\pi}{2} = -\frac{15\pi}{2}. \) Having \( a = 11\pi \) and \( d = -\frac{15\pi}{2} \), we find the sixth term \( a_6 \): \[ a_6 = a + 5d = 11\pi + 5\left(-\frac{15\pi}{2}\right) = 11\pi - \frac{75\pi}{2} = 11\pi - 37.5\pi = -26.5\pi. \] The sixth term of the series is therefore \( -26.5\pi \).

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