2. Given the quadratic pattem: $$ -9 ;-6 ; 1 ; x ; 27 ; \ldots $$ 2.1 Calculate the value of $$ x $$. 2.2 Determine the general term of the pattern. 2.3 Determine $$ T_{100} $$ - 2.4 Which term in the sequence will be equal to 397 ?

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**Problem 2.1:**  
*Calculate the value of $$ x $$ in the quadratic pattern: $$ -9, -6, 1, x, 27, \ldots $$*

To determine $$ x $$, we'll first find the general formula for the $$ n $$-th term of the quadratic sequence. A quadratic sequence can be expressed as:
$$
T_n = an^2 + bn + c
$$
where $$ a $$, $$ b $$, and $$ c $$ are constants.

Given the first few terms:
$$
\begin{align*}
T_1 &= -9 \
T_2 &= -6 \
T_3 &= 1 \
T_4 &= x \
T_5 &= 27
\end{align*}
$$

We can set up the following system of equations using $$ n = 1, 2, 3 $$:

1. For $$ n = 1 $$:
$$
a(1)^2 + b(1) + c = -9 \quad \Rightarrow \quad a + b + c = -9
$$

2. For $$ n = 2 $$:
$$
a(2)^2 + b(2) + c = -6 \quad \Rightarrow \quad 4a + 2b + c = -6
$$

3. For $$ n = 3 $$:
$$
a(3)^2 + b(3) + c = 1 \quad \Rightarrow \quad 9a + 3b + c = 1
$$

**Solving the System:**

Subtract the first equation from the second:
$$
(4a + 2b + c) - (a + b + c) = -6 - (-9) \quad \Rightarrow \quad 3a + b = 3 \quad \text{(Equation 4)}
$$

Subtract the second equation from the third:
$$
(9a + 3b + c) - (4a + 2b + c) = 1 - (-6) \quad \Rightarrow \quad 5a + b = 7 \quad \text{(Equation 5)}
$$

Subtract Equation 4 from Equation 5:
$$
(5a + b) - (3a + b) = 7 - 3 \quad \Rightarrow \quad 2a = 4 \quad \Rightarrow \quad a = 2
$$

Substitute $$ a = 2 $$ into Equation 4:
$$
3(2) + b = 3 \quad \Rightarrow \quad 6 + b = 3 \quad \Rightarrow \quad b = -3
$$

Substitute $$ a = 2 $$ and $$ b = -3 $$ into the first equation:
$$
2 - 3 + c = -9 \quad \Rightarrow \quad c = -8
$$

**General Term Formula:**
$$
T_n = 2n^2 - 3n - 8
$$

**Finding $$ x = T_4 $$:**
$$
T_4 = 2(4)^2 - 3(4) - 8 = 2(16) - 12 - 8 = 32 - 12 - 8 = 12
$$

**Answer:**  
Problem 2.1 Answer:

The value of $$ x $$ is 12.
**Problem 2.1:** The value of $$ x $$ is **12**. **Problem 2.2:** The general term of the pattern is $$ T_n = 2n^2 - 3n - 8 $$. **Problem 2.3:** $$ T_{100} = 2(100)^2 - 3(100) - 8 = 20,000 - 300 - 8 = 19,692 $$. **Problem 2.4:** The term equal to 397 is the 20th term. **Summary:** - $$ x = 12 $$ - General term: $$ T_n = 2n^2 - 3n - 8 $$ - $$ T_{100} = 19,692 $$ - The term equal to 397 is the 20th term.

2. Given the quadratic pattem: \( -9 ;-6 ; 1 ; x ; 27 ; \ldots \) 2.1 Calculate the value of \( x \). 2.2 Determine the general term of the pattern. 2.3 Determine \( T_{100} \) - 2.4 Which term in the sequence will be equal to 397 ?

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