11. The sum of the first five terms of an arithmetic series is zero. The last term is 4 . Determine the first term and the common difference. 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 both problems step by step. --- ### **Problem 11** **Statement:** The sum of the first five terms of an arithmetic series is zero. The last term is 4. Determine the first term and the common difference. **Solution:** Let's denote: - First term of the arithmetic series as \( a \). - Common difference as \( d \). **Step 1: Express the First Five Terms** The first five terms of the arithmetic series are: \[ a, \quad a + d, \quad a + 2d, \quad a + 3d, \quad a + 4d \] **Step 2: Sum of the First Five Terms** Given that the sum of these five terms is zero: \[ S_5 = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 0 \] Simplifying: \[ 5a + 10d = 0 \] \[ 5a = -10d \] \[ a = -2d \quad \text{(Equation 1)} \] **Step 3: Last Term Condition** The last (5th) term is 4: \[ a + 4d = 4 \quad \text{(Given)} \] **Step 4: Substitute Equation 1 into Last Term Condition** \[ -2d + 4d = 4 \] \[ 2d = 4 \] \[ d = 2 \] **Step 5: Find the First Term \( a \)** Using Equation 1: \[ a = -2 \times 2 = -4 \] **Answer:** - **First Term (\( a \))**: \(-4\) - **Common Difference (\( d \))**: \(2\) --- ### **Problem 12** **Statement:** 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. **Solution:** Let's denote: - First term of the arithmetic series as \( a \). - Common difference as \( d \). **Step 1: Sum of the First Five Terms** The first five terms are: \[ a, \quad a + d, \quad a + 2d, \quad a + 3d, \quad a + 4d \] Their sum: \[ S_5 = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d \] Given: \[ 5a + 10d = 5 \] Simplifying: \[ a + 2d = 1 \quad \text{(Equation 1)} \] **Step 2: Relationship Between the Fourth and Seventh Terms** The fourth term: \[ T_4 = a + 3d \] The seventh term: \[ T_7 = a + 6d \] Given that the fourth term exceeds the seventh by 6: \[ T_4 - T_7 = 6 \] \[ (a + 3d) - (a + 6d) = 6 \] \[ -3d = 6 \] \[ d = -2 \] **Step 3: Find the First Term \( a \)** Using Equation 1: \[ a + 2(-2) = 1 \] \[ a - 4 = 1 \] \[ a = 5 \] **Step 4: Sum of the First Ten Terms** The formula for the sum of the first \( n \) terms of an arithmetic series is: \[ S_n = \frac{n}{2} \left[2a + (n - 1)d\right] \] For \( n = 10 \): \[ S_{10} = \frac{10}{2} \left[2 \times 5 + (10 - 1)(-2)\right] \] \[ S_{10} = 5 \left[10 + 9 \times (-2)\right] \] \[ S_{10} = 5 \left[10 - 18\right] \] \[ S_{10} = 5 \times (-8) = -40 \] **Answer:** - **First Term (\( a \))**: \(5\) - **Common Difference (\( d \))**: \(-2\) - **Sum of the First Ten Terms (\( S_{10} \))**: \(-40\) --- **Summary of Answers:** 11. - **First Term**: \(-4\) - **Common Difference**: \(2\) 12. - **First Term**: \(5\) - **Common Difference**: \(-2\) - **Sum of the First Ten Terms**: \(-40\)