2. It can be expanded as, $$ \begin{aligned} \sum_{k=3}^{5} 3 k=3(3) & +3(4)+3(5)=9+12+15 \ & =36 \end{aligned} $$ Copyright reserved Telematics Mathematics Grade 12 Learner Material 5 March to October 2021 Exercises: 1. Given the sequence $$ 6 ; 13 ; 20 ; \ldots $$ a) Determine a formula for the $$ n^{\text {th }} $$ term of the sequence. b) Calculate the $$ 21^{\text {st }} $$ term of this sequence. c) Determine which term of this sequence is 97 . 2. Consider this number pattern: $$ 8 ; 5 ; 2 ; \ldots $$ a) Calculate the $$ 15^{\text {th }} $$ term. b) Determine which term of this sequence is -289 . 3. a) Given the arithmetic sequence $$ 1-p ; 2 p-3 ; p+5 ; \ldots $$ determine the value of $$ p $$. b) Determine the values of the first three terms of the sequence.

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Certainly! Let's work through each of the exercises step by step.

---

## **Exercise 1**

**Given the sequence:** $$ 6 ; 13 ; 20 ; \ldots $$

### **a) Determine a formula for the $$ n^{\text{th}} $$ term of the sequence.**

**Solution:**

This is an **arithmetic sequence** since the difference between consecutive terms is constant.

1. **Find the common difference ($$ d $$)**:
   $$
   d = 13 - 6 = 7 \quad \text{and} \quad 20 - 13 = 7
   $$
   So, $$ d = 7 $$.

2. **Identify the first term ($$ a_1 $$)**:
   $$
   a_1 = 6
   $$

3. **Use the general formula for the $$ n^{\text{th}} $$ term of an arithmetic sequence**:
   $$
   a_n = a_1 + (n - 1)d
   $$
   Plugging in the known values:
   $$
   a_n = 6 + (n - 1) \times 7 = 6 + 7n - 7 = 7n - 1
   $$

**Answer:**
$$
a_n = 7n - 1
$$

---

### **b) Calculate the $$ 21^{\text{st}} $$ term of this sequence.**

**Solution:**

Using the formula derived above:
$$
a_{21} = 7 \times 21 - 1 = 147 - 1 = 146
$$

**Answer:**
$$
a_{21} = 146
$$

---

### **c) Determine which term of this sequence is 97.**

**Solution:**

We need to find $$ n $$ such that $$ a_n = 97 $$:
$$
7n - 1 = 97
$$
$$
7n = 97 + 1 = 98
$$
$$
n = \frac{98}{7} = 14
$$

**Answer:**
The $$ 14^{\text{th}} $$ term is 97.

---

## **Exercise 2**

**Consider this number pattern:** $$ 8 ; 5 ; 2 ; \ldots $$

### **a) Calculate the $$ 15^{\text{th}} $$ term.**

**Solution:**

First, determine if this is an arithmetic sequence.

1. **Find the common difference ($$ d $$)**:
   $$
   d = 5 - 8 = -3 \quad \text{and} \quad 2 - 5 = -3
   $$
   So, $$ d = -3 $$.

2. **Identify the first term ($$ a_1 $$)**:
   $$
   a_1 = 8
   $$

3. **Use the general formula for the $$ n^{\text{th}} $$ term**:
   $$
   a_n = a_1 + (n - 1)d = 8 + (n - 1)(-3)
   $$
   For $$ n = 15 $$:
   $$
   a_{15} = 8 + (15 - 1)(-3) = 8 + 14 \times (-3) = 8 - 42 = -34
   $$

**Answer:**
$$
a_{15} = -34
$$

---

### **b) Determine which term of this sequence is -289.**

**Solution:**

Find $$ n $$ such that $$ a_n = -289 $$.

Use the formula:
$$
a_n = 8 + (n - 1)(-3) = 8 - 3(n - 1)
$$
Set $$ a_n = -289 $$:
$$
8 - 3(n - 1) = -289
$$
$$
-3(n - 1) = -289 - 8 = -297
$$
$$
n - 1 = \frac{-297}{-3} = 99
$$
$$
n = 99 + 1 = 100
$$

**Answer:**
$$
a_{100} = -289
$$
So, -289 is the $$ 100^{\text{th}} $$ term.

---

## **Exercise 3**

### **a) Given the arithmetic sequence $$ 1 - p ; 2p - 3 ; p + 5 ; \ldots $$, determine the value of $$ p $$.**

**Solution:**

In an arithmetic sequence, the difference between consecutive terms is constant.

Let’s denote the terms:

$$
a_1 = 1 - p \
a_2 = 2p - 3 \
a_3 = p + 5
$$

1. **Find the common difference ($$ d $$)**:
   $$
   d = a_2 - a_1 = (2p - 3) - (1 - p) = 2p - 3 -1 + p = 3p -4
   $$
   Similarly,
   $$
   d = a_3 - a_2 = (p + 5) - (2p - 3) = p + 5 -2p +3 = -p +8
   $$
   
2. **Set the two expressions for $$ d $$ equal to each other**:
   $$
   3p - 4 = -p + 8
   $$
   $$
   3p + p = 8 + 4
   $$
   $$
   4p = 12
   $$
   $$
   p = 3
   $$

**Answer:**
$$
p = 3
$$

---

### **b) Determine the values of the first three terms of the sequence.**

**Solution:**

Now that we know $$ p = 3 $$, substitute back into each term:

1. **First term ($$ a_1 $$)**:
   $$
   a_1 = 1 - p = 1 - 3 = -2
   $$

2. **Second term ($$ a_2 $$)**:
   $$
   a_2 = 2p - 3 = 2 \times 3 - 3 = 6 - 3 = 3
   $$

3. **Third term ($$ a_3 $$)**:
   $$
   a_3 = p + 5 = 3 + 5 = 8
   $$

**Answer:**
$$
\text{First three terms: } -2 ; 3 ; 8
$$

---

Feel free to ask if you need further clarification on any of these problems!
**Exercise 1:** a) The $$ n^{\text{th}} $$ term is $$ 7n - 1 $$. b) The $$ 21^{\text{st}} $$ term is 146. c) The $$ 14^{\text{th}} $$ term is 97. **Exercise 2:** a) The $$ 15^{\text{th}} $$ term is -34. b) -289 is the $$ 100^{\text{th}} $$ term. **Exercise 3:** a) $$ p = 3 $$. b) The first three terms are -2, 3, and 8.

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