1. **State the problem:** We have a sequence defined by $u_1 = 6$ and the recursive formula $u_{n+1} = k u_n + 3$, where $k$ is a positive constant. 2. **Find an expression for $u_3$ in terms of $k$:** - First, find $u_2$ using the formula: $$u_2 = k u_1 + 3 = k \times 6 + 3 = 6k + 3$$ - Next, find $u_3$: $$u_3 = k u_2 + 3 = k(6k + 3) + 3 = 6k^2 + 3k + 3$$ 3. **Use the sum condition to find $k$:** - The sum of the first three terms is given as: $$\sum_{n=1}^3 u_n = u_1 + u_2 + u_3 = 117$$ - Substitute the expressions: $$6 + (6k + 3) + (6k^2 + 3k + 3) = 117$$ - Combine like terms: $$6k^2 + 3k + 6k + 3 + 3 + 6 = 117$$ $$6k^2 + 9k + 12 = 117$$ - Simplify: $$6k^2 + 9k + 12 - 117 = 0$$ $$6k^2 + 9k - 105 = 0$$ - Divide entire equation by 3 to simplify: $$\cancel{3} \times 2k^2 + \cancel{3} \times 3k - \cancel{3} \times 35 = 0$$ $$2k^2 + 3k - 35 = 0$$ 4. **Solve the quadratic equation:** - Use the quadratic formula: $$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=3$, $c=-35$. - Calculate the discriminant: $$\Delta = 3^2 - 4 \times 2 \times (-35) = 9 + 280 = 289$$ - Find the roots: $$k = \frac{-3 \pm \sqrt{289}}{2 \times 2} = \frac{-3 \pm 17}{4}$$ - Possible values: $$k_1 = \frac{-3 + 17}{4} = \frac{14}{4} = 3.5$$ $$k_2 = \frac{-3 - 17}{4} = \frac{-20}{4} = -5$$ 5. **Choose the positive value:** Since $k$ is positive, we take: $$k = 3.5$$ **Final answers:** - Expression for $u_3$: $$u_3 = 6k^2 + 3k + 3$$ - Value of $k$: $$k = 3.5$$