3. Determine the smallest number of terms for which... \( \sum_{r=1}^{k}(3 r-2)>117 \)

To determine the smallest number of terms \( k \) for which the sum \[ \sum_{r=1}^{k}(3r - 2) > 117 \] we can follow these steps: ### Step 1: Simplify the Sum First, express the sum in a simpler form: \[ \sum_{r=1}^{k}(3r - 2) = 3\sum_{r=1}^{k} r - 2\sum_{r=1}^{k} 1 \] \[ = 3\left(\frac{k(k + 1)}{2}\right) - 2k \] \[ = \frac{3k(k + 1)}{2} - 2k \] \[ = \frac{3k^2 + 3k - 4k}{2} \] \[ = \frac{3k^2 - k}{2} \] ### Step 2: Set Up the Inequality Set the sum greater than 117: \[ \frac{3k^2 - k}{2} > 117 \] Multiply both sides by 2 to eliminate the denominator: \[ 3k^2 - k > 234 \] ### Step 3: Rearrange the Inequality Bring all terms to one side to form a quadratic inequality: \[ 3k^2 - k - 234 > 0 \] ### Step 4: Solve the Quadratic Equation First, solve the equation \( 3k^2 - k - 234 = 0 \) to find the critical points. Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 3, \quad b = -1, \quad c = -234 \] \[ k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-234)}}{2 \cdot 3} \] \[ k = \frac{1 \pm \sqrt{1 + 2808}}{6} \] \[ k = \frac{1 \pm \sqrt{2809}}{6} \] \[ \sqrt{2809} = 53 \] \[ k = \frac{1 \pm 53}{6} \] This gives two solutions: \[ k = \frac{54}{6} = 9 \quad \text{and} \quad k = \frac{-52}{6} \approx -8.67 \] Since \( k \) must be a positive integer, we consider \( k = 9 \). ### Step 5: Determine the Smallest \( k \) Check if \( k = 9 \) satisfies the original inequality: \[ S(9) = \frac{3(9)^2 - 9}{2} = \frac{243 - 9}{2} = \frac{234}{2} = 117 \] Since \( S(9) = 117 \) is **not** greater than 117, we need to check the next integer, \( k = 10 \): \[ S(10) = \frac{3(10)^2 - 10}{2} = \frac{300 - 10}{2} = \frac{290}{2} = 145 \] \[ 145 > 117 \] ### Conclusion The smallest number of terms \( k \) for which \( \sum_{r=1}^{k}(3r - 2) > 117 \) is **10**. **Answer:** \( k = 10 \)