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Let's analyze the quadratic pattern given in the sequence: \(3, -1, 2, \dots\).

### Step 1: Find the second differences
Since the pattern is quadratic, the second differences should be constant.

1. First term: \( 3 \)
2. Second term: \( -1 \)
3. Third term: \( 2 \)

Now, calculate the first differences:
- Between the first and second terms: \( -1 - 3 = -4 \)
- Between the second and third terms: \( 2 - (-1) = 3 \)

Next, calculate the second difference:
- Between \( -4 \) and \( 3 \): \( 3 - (-4) = 7 \)

Thus, the second difference is constant and equal to 7, which is consistent with a quadratic sequence.

### Step 2: General form of a quadratic sequence
The general form for the \(n\)-th term of a quadratic sequence is:
\[
T_n = an^2 + bn + c
\]
We need to find the values of \(a\), \(b\), and \(c\) by using the first three terms of the sequence.

#### Substituting the first three terms:

1. For \(n = 1\), \(T_1 = 3\):
\[
a(1)^2 + b(1) + c = 3 \implies a + b + c = 3
\]
2. For \(n = 2\), \(T_2 = -1\):
\[
a(2)^2 + b(2) + c = -1 \implies 4a + 2b + c = -1
\]
3. For \(n = 3\), \(T_3 = 2\):
\[
a(3)^2 + b(3) + c = 2 \implies 9a + 3b + c = 2
\]

Now we solve this system of equations.

### Step 3: Solving the system of equations
We have the following system:
1. \( a + b + c = 3 \)
2. \( 4a + 2b + c = -1 \)
3. \( 9a + 3b + c = 2 \)

Let's subtract equation 1 from equation 2:
\[
(4a + 2b + c) - (a + b + c) = -1 - 3 \implies 3a + b = -4 \tag{4}
\]

Now subtract equation 2 from equation 3:
\[
(9a + 3b + c) - (4a + 2b + c) = 2 - (-1) \implies 5a + b = 3 \tag{5}
\]

Now subtract equation (4) from equation (5):
\[
(5a + b) - (3a + b) = 3 - (-4) \implies 2a = 7 \implies a = \frac{7}{2}
\]

Substitute \(a = \frac{7}{2}\) into equation (4):
\[
3\left(\frac{7}{2}\right) + b = -4 \implies \frac{21}{2} + b = -4 \implies b = -4 - \frac{21}{2} = -\frac{29}{2}
\]

Now substitute \(a = \frac{7}{2}\) and \(b = -\frac{29}{2}\) into equation 1:
\[
\frac{7}{2} - \frac{29}{2} + c = 3 \implies \frac{-22}{2} + c = 3 \implies -11 + c = 3 \implies c = 14
\]

### Step 4: General term
Thus, the general term is:
\[
T_n = \frac{7}{2}n^2 - \frac{29}{2}n + 14
\]

### Step 5: Find which term equals 1839
We are asked to find which term equals 1839. Set \(T_n = 1839\):
\[
\frac{7}{2}n^2 - \frac{29}{2}n + 14 = 1839
\]

Multiply the entire equation by 2 to eliminate the fractions:
\[
7n^2 - 29n + 28 = 3678
\]

Simplify:
\[
7n^2 - 29n - 3650 = 0
\]

Solve this quadratic equation using the quadratic formula:
\[
n = \frac{-(-29) \pm \sqrt{(-29)^2 - 4(7)(-3650)}}{2(7)}
\]
\[
n = \frac{29 \pm \sqrt{841 + 102200}}{14}
\]
\[
n = \frac{29 \pm \sqrt{103041}}{14}
\]
\[
n = \frac{29 \pm 321}{14}
\]

Thus:
1. \( n = \frac{29 + 321}{14} = \frac{350}{14} = 25 \)
2. \( n = \frac{29 - 321}{14} = \frac{-292}{14} = -20.86 \) (not a valid solution)

Therefore, the term that equals 1839 is the 25th term.

---

Would you like further clarification or details? Here are some related questions you might want to explore:

1. How can you solve a quadratic sequence using different methods?
2. What are the applications of quadratic sequences in real-life scenarios?
3. How does the second difference relate to the general formula of a quadratic sequence?
4. Could this pattern be modeled in other ways, like using recursive formulas?
5. How would changing the coefficients affect the general term?

**Tip:** When solving quadratic equations, always check the discriminant (\(b^2 - 4ac\)) to ensure the solutions are real and applicable.

Consider the quadratic pattern: 3, -1, 2, ... Determine the general term (T_n). Which term of the pattern is equal to 1839?

This math problem explores finding the general term of a quadratic sequence given the terms 3, -1, 2... It involves calculating the second differences and solving systems of equations. Additionally, it shows how to determine which term equals 1839 using the quadratic formula. Suitable for Grades 9-11.

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