1. **State the problem:** We are given the sequence 4, 7, 12, 19, 28, 39 and need to find a formula for the nth term, $a_n$. 2. **Analyze the sequence:** Calculate the differences between terms: $$7-4=3, \quad 12-7=5, \quad 19-12=7, \quad 28-19=9, \quad 39-28=11$$ These are the first differences: 3, 5, 7, 9, 11. 3. **Check second differences:** Calculate differences of first differences: $$5-3=2, \quad 7-5=2, \quad 9-7=2, \quad 11-9=2$$ The second differences are constant and equal to 2. 4. **Interpretation:** Constant second differences indicate the sequence is quadratic, so $a_n$ has the form: $$a_n = An^2 + Bn + C$$ 5. **Set up equations using known terms:** For $n=1$, $a_1=4$: $$A(1)^2 + B(1) + C = 4 \Rightarrow A + B + C = 4$$ For $n=2$, $a_2=7$: $$4A + 2B + C = 7$$ For $n=3$, $a_3=12$: $$9A + 3B + C = 12$$ 6. **Solve the system:** From equations: $$\begin{cases} A + B + C = 4 \\ 4A + 2B + C = 7 \\ 9A + 3B + C = 12 \end{cases}$$ Subtract first from second: $$\cancel{4A} + 2B + C - (\cancel{A} + B + C) = 7 - 4 \Rightarrow 3A + B = 3$$ Subtract second from third: $$9A + 3B + C - (4A + 2B + C) = 12 - 7 \Rightarrow 5A + B = 5$$ Subtract the two new equations: $$(5A + B) - (3A + B) = 5 - 3 \Rightarrow 2A = 2 \Rightarrow A = 1$$ Plug $A=1$ into $3A + B = 3$: $$3(1) + B = 3 \Rightarrow B = 0$$ Plug $A=1$, $B=0$ into $A + B + C = 4$: $$1 + 0 + C = 4 \Rightarrow C = 3$$ 7. **Final formula:** $$a_n = n^2 + 3$$ 8. **Verification:** For $n=4$, $$a_4 = 4^2 + 3 = 16 + 3 = 19$$ Matches the sequence. **Answer:** $$\boxed{a_n = n^2 + 3}$$