1. **State the problem:** We are given the quadratic sequence: -5, -2, 3, 10, 19, ... and need to find the formula for the $n^{th}$ term. 2. **Recall the formula for a quadratic sequence:** The $n^{th}$ term of a quadratic sequence is generally given by $$a n^2 + b n + c$$ where $a$, $b$, and $c$ are constants to be determined. 3. **Find the first differences:** Subtract consecutive terms: $$-2 - (-5) = 3$$ $$3 - (-2) = 5$$ $$10 - 3 = 7$$ $$19 - 10 = 9$$ So the first differences are: 3, 5, 7, 9. 4. **Find the second differences:** Subtract consecutive first differences: $$5 - 3 = 2$$ $$7 - 5 = 2$$ $$9 - 7 = 2$$ The second differences are constant and equal to 2. 5. **Use the second difference to find $a$:** The second difference equals $2a$, so $$2a = 2 \implies a = 1$$ 6. **Set up equations to find $b$ and $c$:** Using the formula $T_n = n^2 + b n + c$, plug in the first terms: - For $n=1$, $T_1 = 1^2 + b(1) + c = 1 + b + c = -5$ - For $n=2$, $T_2 = 2^2 + b(2) + c = 4 + 2b + c = -2$ 7. **Solve the system:** From $n=1$: $$1 + b + c = -5 \implies b + c = -6$$ From $n=2$: $$4 + 2b + c = -2 \implies 2b + c = -6$$ Subtract the first from the second: $$(2b + c) - (b + c) = -6 - (-6) \implies b = 0$$ Substitute $b=0$ into $b + c = -6$: $$0 + c = -6 \implies c = -6$$ 8. **Write the $n^{th}$ term formula:** $$T_n = n^2 - 6$$ 9. **Check with a term:** For $n=3$, $$T_3 = 3^2 - 6 = 9 - 6 = 3$$ which matches the sequence. **Final answer:** The $n^{th}$ term rule is $$\boxed{T_n = n^2 - 6}$$