12. The sum of the first five terms of an arithmetic series is 5. The fourth term exceeds the seventh by 6. Find the sum of the first ten terms.

Let's solve the problem step by step. **Given:** - The series is arithmetic with first term \( a \) and common difference \( d \). - The sum of the first five terms (\( S_5 \)) is 5. - The fourth term exceeds the seventh by 6. **1. Sum of the first five terms:** The sum of the first \( n \) terms of an arithmetic series is given by: \[ S_n = \frac{n}{2} \left[2a + (n-1)d\right] \] For \( S_5 = 5 \): \[ \frac{5}{2} \left[2a + 4d\right] = 5 \\ 2a + 4d = 2 \quad \Rightarrow \quad a + 2d = 1 \quad \text{(Equation 1)} \] **2. Relation between the fourth and seventh terms:** The \( n \)-th term of an arithmetic series is: \[ a_n = a + (n-1)d \] Given that the fourth term exceeds the seventh by 6: \[ (a + 3d) - (a + 6d) = 6 \\ -3d = 6 \quad \Rightarrow \quad d = -2 \] **3. Finding the first term \( a \):** Substitute \( d = -2 \) into Equation 1: \[ a + 2(-2) = 1 \\ a - 4 = 1 \\ a = 5 \] **4. Sum of the first ten terms:** Using the sum formula for \( n = 10 \): \[ S_{10} = \frac{10}{2} \left[2(5) + 9(-2)\right] \\ S_{10} = 5 \left[10 - 18\right] \\ S_{10} = 5 \times (-8) \\ S_{10} = -40 \] **Answer:** -40