1.1 Prove that the sum of $$ n $$ terms of an arithmetic series is given by the following formula: 1.2 Given the arithmetic series $$ 5+9+13+\ldots $$ 1.2.1 Determine the sum of the first $$ n $$ terms of the series in terms of $$ n $$. Leave your answer in simplified form. 1.2.2 Calculate the value of the $$ 29^{ ext {th }} $$ term of the series. 1.2 .3 Calculate the value of $$ n $$ if it is further given that the sum of the first $$ n $$ terms of the series is 2015 .

Question

1.1 Prove that the sum of $$ n $$ terms of an arithmetic series is given by the following formula: 1.2 Given the arithmetic series $$ 5+9+13+\ldots $$ 1.2.1 Determine the sum of the first $$ n $$ terms of the series in terms of $$ n $$. Leave your answer in simplified form. 1.2.2 Calculate the value of the $$ 29^{	ext {th }} $$ term of the series. 1.2 .3 Calculate the value of $$ n $$ if it is further given that the sum of the first $$ n $$ terms of the series is 2015 .

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### **1.1 Prove the Sum Formula for an Arithmetic Series**

**The Formula:**
The sum of the first $$ n $$ terms of an arithmetic series is given by:
$$
S_n = \frac{n}{2} \left(2a + (n - 1)d\right)
$$
where:
- $$ a $$ is the first term,
- $$ d $$ is the common difference,
- $$ n $$ is the number of terms.

**Proof:**
An arithmetic series is a sequence where each term increases by a constant difference $$ d $$. Consider the series:
$$
a, \, a + d, \, a + 2d, \, \ldots, \, a + (n-1)d
$$

The sum of the first $$ n $$ terms can be written as:
$$
S_n = a + (a + d) + (a + 2d) + \ldots + [a + (n-1)d]
$$

Writing the sum in reverse and adding both equations:
$$
S_n = a + (a + d) + (a + 2d) + \ldots + [a + (n-1)d]
$$
$$
S_n = [a + (n-1)d] + [a + (n-2)d] + \ldots + a
$$
$$
2S_n = n \cdot 2a + d \cdot n(n-1)
$$
$$
2S_n = n(2a + (n-1)d)
$$
$$
S_n = \frac{n}{2} (2a + (n-1)d)
$$
This proves the formula.

---

### **1.2 Given the Arithmetic Series $$ 5 + 9 + 13 + \ldots $$**

**Given:**
- First term ($$ a $$) = 5
- Common difference ($$ d $$) = 9 - 5 = 4

#### **1.2.1 Determine the Sum of the First $$ n $$ Terms**

Using the sum formula:
$$
S_n = \frac{n}{2} \left(2a + (n - 1)d\right)
$$
Plugging in the values:
$$
S_n = \frac{n}{2} \left(2 \times 5 + (n - 1) \times 4\right)
$$
$$
S_n = \frac{n}{2} (10 + 4n - 4)
$$
$$
S_n = \frac{n}{2} (4n + 6)
$$
$$
S_n = n(2n + 3)
$$
$$
\boxed{S_n = 2n^2 + 3n}
$$
**Simplified Form:** $$ S_n = 2n^2 + 3n $$

#### **1.2.2 Calculate the Value of the $$ 29^{\text{th}} $$ Term**

The $$ n^{\text{th}} $$ term ($$ a_n $$) of an arithmetic series is:
$$
a_n = a + (n - 1)d
$$
For $$ n = 29 $$:
$$
a_{29} = 5 + (29 - 1) \times 4
$$
$$
a_{29} = 5 + 28 \times 4
$$
$$
a_{29} = 5 + 112
$$
$$
\boxed{a_{29} = 117}
$$

#### **1.2.3 Calculate $$ n $$ When $$ S_n = 2015 $$**

We have the sum formula:
$$
S_n = 2n^2 + 3n = 2015
$$
Rearrange to form a quadratic equation:
$$
2n^2 + 3n - 2015 = 0
$$

**Solving the Quadratic Equation:**
$$
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Where $$ a = 2 $$, $$ b = 3 $$, and $$ c = -2015 $$:
$$
n = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2015)}}{2 \times 2}
$$
$$
n = \frac{-3 \pm \sqrt{9 + 16120}}{4}
$$
$$
n = \frac{-3 \pm \sqrt{16129}}{4}
$$
$$
\sqrt{16129} = 127
$$
$$
n = \frac{-3 \pm 127}{4}
$$

**Possible Solutions:**
1. $$ n = \frac{-3 + 127}{4} = \frac{124}{4} = 31 $$
2. $$ n = \frac{-3 - 127}{4} = \frac{-130}{4} = -32.5 $$ (Discarded, since $$ n $$ must be positive)

$$
\boxed{n = 31}
$$

---

**Summary of Answers:**
1. **Sum of first $$ n $$ terms:** $$ S_n = 2n^2 + 3n $$
2. **29th term:** $$ a_{29} = 117 $$
3. **Number of terms for $$ S_n = 2015 $$:** $$ n = 31 $$
**1.1 Sum Formula:** The sum of the first $$ n $$ terms of an arithmetic series is $$ S_n = 2n^2 + 3n $$. **1.2.1 Sum of the First $$ n $$ Terms:** $$ S_n = 2n^2 + 3n $$ **1.2.2 29th Term:** $$ a_{29} = 117 $$ **1.2.3 Number of Terms for Sum 2015:** $$ n = 31 $$

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