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This problem involves finding the general \( n \)-th term of a quadratic sequence. Let’s solve it step by step.

### Sequence: \(-6, -3, 2, 9, 18, \dots\)

#### Step 1: Find the first differences.
\[
\text{First differences: } (-3 - (-6)), (2 - (-3)), (9 - 2), (18 - 9)
\]
\[
= 3, 5, 7, 9
\]

#### Step 2: Find the second differences.
\[
\text{Second differences: } (5 - 3), (7 - 5), (9 - 7)
\]
\[
= 2, 2, 2
\]

Since the second differences are constant, this confirms the sequence is quadratic.

#### Step 3: General form of a quadratic sequence.
The \( n \)-th term of a quadratic sequence is given by:
\[
T_n = an^2 + bn + c
\]
We need to determine the coefficients \( a \), \( b \), and \( c \) by substituting values for \( n \) and \( T_n \).

#### Step 4: Solve for \( a \), \( b \), and \( c \).
Using the terms:
- For \( n = 1, T_1 = -6 \):
\[
a(1)^2 + b(1) + c = -6 \quad \Rightarrow \quad a + b + c = -6
\]

- For \( n = 2, T_2 = -3 \):
\[
a(2)^2 + b(2) + c = -3 \quad \Rightarrow \quad 4a + 2b + c = -3
\]

- For \( n = 3, T_3 = 2 \):
\[
a(3)^2 + b(3) + c = 2 \quad \Rightarrow \quad 9a + 3b + c = 2
\]

This gives us the system of equations:
1. \( a + b + c = -6 \)
2. \( 4a + 2b + c = -3 \)
3. \( 9a + 3b + c = 2 \)

#### Step 5: Solve the system of equations.
- Subtract (1) from (2):
\[
(4a + 2b + c) - (a + b + c) = -3 - (-6)
\]
\[
3a + b = 3 \quad \text{(Equation 4)}
\]

- Subtract (2) from (3):
\[
(9a + 3b + c) - (4a + 2b + c) = 2 - (-3)
\]
\[
5a + b = 5 \quad \text{(Equation 5)}
\]

- Subtract (4) from (5):
\[
(5a + b) - (3a + b) = 5 - 3
\]
\[
2a = 2 \quad \Rightarrow \quad a = 1
\]

- Substitute \( a = 1 \) into (4):
\[
3(1) + b = 3 \quad \Rightarrow \quad b = 0
\]

- Substitute \( a = 1 \) and \( b = 0 \) into (1):
\[
1 + 0 + c = -6 \quad \Rightarrow \quad c = -7
\]

#### Step 6: Write the \( n \)-th term formula.
\[
T_n = an^2 + bn + c \quad \Rightarrow \quad T_n = n^2 - 7
\]

### Final Answer:
The \( n \)-th term rule for the sequence is:
\[
T_n = n^2 - 7
\]

Let me know if you'd like further clarifications! Here are some related questions for practice:

---

1. How would you check if a given sequence is quadratic?
2. Can you determine the \( n \)-th term formula for the sequence \( 2, 6, 12, 20, 30, \dots \)?
3. What happens to the \( n \)-th term formula if the second differences are not constant?
4. How can you verify your answer for the \( n \)-th term using substitution?
5. What is the difference between an arithmetic sequence and a quadratic sequence?

**Tip**: For quadratic sequences, always confirm the second differences are constant before proceeding.

The start of a quadratic sequence is shown below. What is the nth term rule for the sequence? -6, -3, 2, 9, 18, ...

Learn how to find the nth term rule for the quadratic sequence -6, -3, 2, 9, 18. Includes a detailed step-by-step breakdown of solving for a, b, and c in the general quadratic formula Tn = an^2 + bn + c. Suitable for Grades 8-10 students studying algebra.

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