Determine the sum to 50 terms of the arithmetic series of which the 7 th term is 33 and the 15 th term is 65 . In an arithmetic series, \( T_{3}=-8 \) and \( T_{12}=-35 \). Determine the number of terms in the series if its sum is -100 . In an arithmetic series, \( T_{7}=45 \) and \( T_{3}=17 \). Determine the number of terms in the series if \( S_{n}=123 \). The sum of the first 6 terms of an arithmetic series is 90 and the last term (6th) is 25 . Determine the first term and the common difference.

To find the sum of the first 50 terms of the given arithmetic series, we first need to determine the first term and the common difference. 1. The 7th term (\(T_7\)) is 33, and the 15th term (\(T_{15}\)) is 65. Since the nth term of an arithmetic series can be expressed as: \[ T_n = a + (n-1)d \] Setting up the equations: \[ T_7 = a + 6d = 33 \quad \text{(1)} \] \[ T_{15} = a + 14d = 65 \quad \text{(2)} \] Subtract equation (1) from (2): \[ (a + 14d) - (a + 6d) = 65 - 33 \implies 8d = 32 \implies d = 4 \] Substitute \(d\) back into equation (1): \[ a + 6(4) = 33 \implies a + 24 = 33 \implies a = 9 \] Now we can find the sum \(S_n\) of the first 50 terms: \[ S_{50} = \frac{n}{2}(2a + (n-1)d) = \frac{50}{2}(2(9) + (50-1)(4)) = 25(18 + 196) = 25(214) = 5350 \] The first term is 9, and the common difference is 4. --- For the next arithmetic series, we have \(T_3 = -8\) and \(T_{12} = -35\) with a sum of -100. 1. Using the formulas: \[ T_3 = a + 2d = -8 \quad \text{(1)} \] \[ T_{12} = a + 11d = -35 \quad \text{(2)} \] We can subtract (1) from (2): \[ (a + 11d) - (a + 2d) = -35 + 8 \implies 9d = -27 \implies d = -3 \] Substitute \(d\) back into equation (1) to find \(a\): \[ a + 2(-3) = -8 \implies a - 6 = -8 \implies a = -2 \] Now for the sum \(S_n\): The sum of an arithmetic series is: \[ S_n = \frac{n}{2}(2a + (n - 1)d) \] Setting \(S_n = -100\): \[ -100 = \frac{n}{2}(2(-2) + (n - 1)(-3)) \] Simplifying: \[ -100 = \frac{n}{2}(-4 - 3n + 3) \implies -100 = \frac{n}{2}(-3n - 1) \implies -200 = n(-3n - 1) \] Rearranging gives: \[ 3n^2 + n - 200 = 0 \] Solving this quadratic equation by substituting values or using the quadratic formula, you'll find \(n = 8\) as one of the suitable answers. --- In the scenario with \(T_7 = 45\) and \(T_3 = 17\), we can derive \(d\) and \(a\): 1. From: \[ T_3 = a + 2d = 17 \quad \text{(1)} \] \[ T_7 = a + 6d = 45 \quad \text{(2)} \] Subtracting (1) from (2): \[ (a + 6d) - (a + 2d) = 45 - 17 \implies 4d = 28 \implies d = 7 \] Plugging \(d\) back into (1): \[ a + 2(7) = 17 \implies a + 14 = 17 \implies a = 3 \] With \(S_n = 123\): \[ 123 = \frac{n