2.2 The sum of the first terms in an arithmetic series is given by: $$ S_{n}=n^{2}-2 n $$ Calculate: 2.2 .1 the sum of the first 13 terms. 2.2 .2 the $$ 13^{ ext {th }} $$ term. 2.3 Given the quadratic pattern: $$ x ; 6 ; 9 ; y ; 24 ; \ldots $$ Calculate the sum of $$ x $$ and $$ y $$.

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2.2 The sum of the first terms in an arithmetic series is given by: $$ S_{n}=n^{2}-2 n $$ Calculate: 2.2 .1 the sum of the first 13 terms. 2.2 .2 the $$ 13^{	ext {th }} $$ term. 2.3 Given the quadratic pattern: $$ x ; 6 ; 9 ; y ; 24 ; \ldots $$ Calculate the sum of $$ x $$ and $$ y $$.

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### **2.2 Arithmetic Series with Sum $$ S_{n} = n^{2} - 2n $$**

#### **2.2.1. Sum of the First 13 Terms ($$ S_{13} $$)**

Given:
$$
S_{n} = n^{2} - 2n
$$

To find $$ S_{13} $$:
$$
S_{13} = 13^{2} - 2 \times 13 = 169 - 26 = 143
$$

**Answer:** The sum of the first 13 terms is **143**.

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#### **2.2.2. The 13th Term ($$ a_{13} $$)**

To find the $$ n^{th} $$ term of an arithmetic series when the sum $$ S_{n} $$ is known, use:
$$
a_{n} = S_{n} - S_{n-1}
$$

Calculate $$ a_{13} $$:
$$
S_{13} = 13^{2} - 2 \times 13 = 143 \
S_{12} = 12^{2} - 2 \times 12 = 144 - 24 = 120 \
a_{13} = S_{13} - S_{12} = 143 - 120 = 23
$$

**Answer:** The 13th term is **23**.

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### **2.3 Quadratic Sequence: $$ x ; 6 ; 9 ; y ; 24 ; \ldots $$**

A quadratic sequence has a general term of the form:
$$
a_{n} = an^{2} + bn + c
$$

Given the sequence:
$$
\text{Term 1 (} n=0 \text{)}: x \
\text{Term 2 (} n=1 \text{)}: 6 \
\text{Term 3 (} n=2 \text{)}: 9 \
\text{Term 4 (} n=3 \text{)}: y \
\text{Term 5 (} n=4 \text{)}: 24 \
$$

**Step 1: Establish Equations**

Using $$ n $$ starting from 0:
$$
a_{0}: c = x \
a_{1}: a(1)^2 + b(1) + c = 6 \
a_{2}: a(2)^2 + b(2) + c = 9 \
a_{4}: a(4)^2 + b(4) + c = 24 \
$$

**Step 2: Solve the System of Equations**

1. From $$ a_{1} $$:
$$
a + b + c = 6 \quad \text{(Equation 1)}
$$

2. From $$ a_{2} $$:
$$
4a + 2b + c = 9 \quad \text{(Equation 2)}
$$

3. From $$ a_{4} $$:
$$
16a + 4b + c = 24 \quad \text{(Equation 3)}
$$

**Subtract Equation 1 from Equation 2:**
$$
3a + b = 3 \quad \text{(Equation 4)}
$$

**Subtract Equation 1 from Equation 3:**
$$
15a + 3b = 18 \quad \Rightarrow \quad 5a + b = 6 \quad \text{(Equation 5)}
$$

**Solve Equations 4 and 5:**
$$
\begin{cases}
3a + b = 3 \
5a + b = 6
\end{cases}
$$

**Subtract Equation 4 from Equation 5:**
$$
2a = 3 \quad \Rightarrow \quad a = \frac{3}{2}
$$

**Find $$ b $$ using Equation 4:**
$$
3 \left(\frac{3}{2}\right) + b = 3 \quad \Rightarrow \quad \frac{9}{2} + b = 3 \quad \Rightarrow \quad b = -\frac{3}{2}
$$

**Find $$ c $$ using Equation 1:**
$$
\frac{3}{2} - \frac{3}{2} + c = 6 \quad \Rightarrow \quad c = 6
$$

**Step 3: Find $$ y $$ (4th Term)**
$$
a_{3} = a(3)^2 + b(3) + c = \frac{3}{2} \times 9 - \frac{3}{2} \times 3 + 6 = \frac{27}{2} - \frac{9}{2} + 6 = 9 + 6 = 15
$$

**Step 4: Find $$ x $$ (1st Term)**
$$
x = a_{0} = c = 6
$$

**Step 5: Calculate $$ x + y $$**
$$
x + y = 6 + 15 = 21
$$

**Answer:** The sum of $$ x $$ and $$ y $$ is **21**.

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### **Summary of Answers**
1. **2.2.1:** The sum of the first 13 terms is **143**.
2. **2.2.2:** The 13th term is **23**.
3. **2.3:** The sum of $$ x $$ and $$ y $$ is **21**.
**Answers:** 1. The sum of the first 13 terms is **143**. 2. The 13th term is **23**. 3. The sum of $$ x $$ and $$ y $$ is **21**.

2.2 The sum of the first terms in an arithmetic series is given by: \( S_{n}=n^{2}-2 n \) Calculate: 2.2 .1 the sum of the first 13 terms. 2.2 .2 the \( 13^{\text {th }} \) term. 2.3 Given the quadratic pattern: \( x ; 6 ; 9 ; y ; 24 ; \ldots \) Calculate the sum of \( x \) and \( y \).

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