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.

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 \( -26.5\pi \).