1. **Problem Statement:** Find the formula for the nth term of the sequence: 3, 12, 27, 48, 75, ... 2. **Identify the pattern:** Let's denote the nth term by $a_n$. We want to find a formula for $a_n$. 3. **Check differences:** Calculate the first differences: $$12 - 3 = 9, \quad 27 - 12 = 15, \quad 48 - 27 = 21, \quad 75 - 48 = 27$$ 4. **Check second differences:** $$15 - 9 = 6, \quad 21 - 15 = 6, \quad 27 - 21 = 6$$ Since the second differences are constant (6), the sequence is quadratic, so $a_n$ has the form: $$a_n = An^2 + Bn + C$$ 5. **Set up equations using known terms:** For $n=1$, $a_1 = A(1)^2 + B(1) + C = A + B + C = 3$ For $n=2$, $a_2 = 4A + 2B + C = 12$ For $n=3$, $a_3 = 9A + 3B + C = 27$ 6. **Solve the system:** From the equations: $$\begin{cases} A + B + C = 3 \\ 4A + 2B + C = 12 \\ 9A + 3B + C = 27 \end{cases}$$ Subtract the first from the second: $$ (4A - A) + (2B - B) + (C - C) = 12 - 3 \Rightarrow 3A + B = 9 $$ Subtract the second from the third: $$ (9A - 4A) + (3B - 2B) + (C - C) = 27 - 12 \Rightarrow 5A + B = 15 $$ Subtract the two new equations: $$ (5A + B) - (3A + B) = 15 - 9 \Rightarrow 2A = 6 \Rightarrow A = 3 $$ Plug $A=3$ into $3A + B = 9$: $$3(3) + B = 9 \Rightarrow 9 + B = 9 \Rightarrow B = 0$$ Plug $A=3$, $B=0$ into $A + B + C = 3$: $$3 + 0 + C = 3 \Rightarrow C = 0$$ 7. **Final formula:** $$a_n = 3n^2$$ 8. **Verification:** For $n=1$, $a_1 = 3(1)^2 = 3$ For $n=2$, $a_2 = 3(2)^2 = 12$ For $n=3$, $a_3 = 3(3)^2 = 27$ For $n=4$, $a_4 = 3(4)^2 = 48$ For $n=5$, $a_5 = 3(5)^2 = 75$ All match the given sequence. **Answer:** The nth term formula is: $$a_n = 3n^2$$