(h) Consider the arithmetic series \( \frac{3}{2}+2+\frac{5}{2}+\ldots \) (1) Calculate \( S_{12} \). (2) Calculate the value of \( k \) if \( S_{k}=231 \). (3) Calculate the greatest number of terms that can be added, if the answer must be less than 300 .

To analyze the arithmetic series \( \frac{3}{2}+2+\frac{5}{2}+\ldots \), we first need to identify the first term and the common difference. 1. **Calculate \( S_{12} \)**: The first term \( a \) is \( \frac{3}{2} \) and the common difference \( d \) can be calculated from the first two terms: \[ d = 2 - \frac{3}{2} = \frac{1}{2} \] The formula for the sum of the first \( n \) terms of an arithmetic series is: \[ S_n = \frac{n}{2} (2a + (n - 1)d) \] Plugging in the values for \( n = 12 \): \[ S_{12} = \frac{12}{2} \left(2 \cdot \frac{3}{2} + (12 - 1) \cdot \frac{1}{2}\right) = 6 \left(3 + 5.5\right) = 6 \cdot 8.5 = 51 \] Thus, \( S_{12} = 51 \). 2. **Calculate the value of \( k \) if \( S_{k}=231 \)**: We can set up the equation: \[ S_k = \frac{k}{2} (2a + (k - 1)d) = 231 \] Substituting \( a = \frac{3}{2} \) and \( d = \frac{1}{2} \): \[ \frac{k}{2} \left(2 \cdot \frac{3}{2} + (k - 1) \cdot \frac{1}{2}\right) = 231 \] Simplifying gives: \[ \frac{k}{2} \left(3 + \frac{k - 1}{2}\right) = 231 \] Multiplying both sides by 2: \[ k \left(3 + \frac{k - 1}{2}\right) = 462 \] Expanding: \[ 6k + k(k - 1) = 924 \] Rearranging gives: \[ k^2 + 5k - 924 = 0 \] Using the quadratic formula: \[ k = \frac{-5 \pm \sqrt{5^2 + 4 \cdot 924}}{2} = \frac{-5 \pm \sqrt{3649}}{2} \] The positive integer solution (since \( k \) must be positive) is approximately \( k = 28 \). 3. **Calculate the greatest number of terms that can be added, if the answer must be less than 300**: We need to solve for \( k \) in: \[ S_k < 300 \] From the sum formula: \[ S_k = \frac{k}{2} \left(3 + \frac{k - 1}{2}\right) \] Setting up the inequality: \[ \frac{k}{2} \left(3 + \frac{k - 1}{2}\right) < 300 \] Multiplying by 2 and simplifying: \[ k \left(3 + \frac{k - 1}{2}\right) < 600 \] Solving similarly as above: \[ k^2 + 5k - 600 < 0 \] The roots are calculated using the quadratic equation: \[ k = \frac{-5 \pm \sqrt{5^2 + 2400}}{2} = \frac{-5 \pm 49}{2} \] The viable root gives \( k = 22 \) (the positive root). Thus, the greatest number of terms that can be summed while keeping the sum under 300 is \( k = 22 \).