1. **State the problem:** We are given the quadratic sequence defined by the formula $a_n = n^2 + 6n - 10$. 2. **List the first 5 terms:** To find the first 5 terms, substitute $n=1,2,3,4,5$ into the formula. - For $n=1$: $1^2 + 6(1) - 10 = 1 + 6 - 10 = -3$ - For $n=2$: $2^2 + 6(2) - 10 = 4 + 12 - 10 = 6$ - For $n=3$: $3^2 + 6(3) - 10 = 9 + 18 - 10 = 17$ - For $n=4$: $4^2 + 6(4) - 10 = 16 + 24 - 10 = 30$ - For $n=5$: $5^2 + 6(5) - 10 = 25 + 30 - 10 = 45$ So the first 5 terms are $-3, 6, 17, 30, 45$. 3. **Check if 765 appears in the sequence:** We want to find if there exists an integer $n$ such that $$n^2 + 6n - 10 = 765$$ Rearranging, $$n^2 + 6n - 775 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula $$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=6$, and $c=-775$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 6^2 - 4(1)(-775) = 36 + 3100 = 3136$$ Since $\sqrt{3136} = 56$, the roots are $$n = \frac{-6 \pm 56}{2}$$ Calculate each root: - $n = \frac{-6 + 56}{2} = \frac{50}{2} = 25$ - $n = \frac{-6 - 56}{2} = \frac{-62}{2} = -31$ 5. **Interpretation:** Since $n$ represents the term number in the sequence, it must be a positive integer. $n=25$ is valid. Therefore, the number 765 does appear in the sequence as the 25th term. **Final answers:** - First 5 terms: $-3, 6, 17, 30, 45$ - 765 appears in the sequence at $n=25$.