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# Arithmetic Sequence

• define an arithmetic sequence;
• find the common difference of an arithmetic series;
• determine the nth term of an arithmetic series;
• formulate a general formula for the nth term of any arithmetic series;
• differentiate arithmetic sequence from arithmetic series;
• solve for the arithmetic series; and
• know the importance of learning arithmetic series.

## What is an arithmetic sequence?

An arithmetic progression or sequence is a collection of numbers in which the difference between consecutive terms is a constant. It is said to be an arithmetic sequence if the sequence is created by adding or subtracting the same number each time. In a nutshell, it is a set of numbers that is in order.

For example, the sequence 3, 18, 13, 18, 23, 28, 33 is an arithmetic progression with a common difference of 5.

The formal definition of an arithmetic sequence, or arithmetic progression, defines it as a sequence of numbers in which each subsequent number is the sum of the preceding number and some constant d.

Mathematically, an arithmetic sequence is in the form

a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …

where,
a is the first term of the sequence; and
d is the common difference of the sequence.

## What are the terms used in an arithmetic sequence?

There are specific terms used in an arithmetic sequence that you need to remember as it will help us solve problems involving arithmetic sequence.

• Terms – in an arithmetic sequence, each number is called as a term. Alternatively, we can also call these numbers as elements or members.

Say, for example, in the sequence of 3, 8, 13, 18, 23, 28, and 33. Each number is named based on their order in the sequence.

• First term – in an arithmetic sequence, the first term, as the name implies, is the initial term in the sequence. It is denoted by a1 or a.

Say, for example, in the sequence of 3, 8, 13, 18, 23, 28, and 33, the first term is 3. When solving problems involving arithmetic sequence, we can denote it as a1 = 3 or a = 3.

• Common difference – the common difference of an arithmetic sequence is the difference between two consecutive terms of an arithmetic progression.

For example, in the sequence of 3, 8, 13, 18, 23, 28, and 33, the common difference is determined by subtracting:

8 – 3 = 5; 13 – 8 = 5; 18 – 13 = 5; 23 – 18 = 5; 28 – 23 = 5;  33 – 38 = 5.

We can show the common difference of an arithmetic sequence as:

or as:

Since the difference of all the two consecutive terms of the sequence is equal, we can say that the common difference is 5.

• nth term – the nth term of an arithmetic sequence is any term of the sequence which is not given in the sequence. The nth term of an arithmetic sequence is defined by the first term and the common difference of the sequence.

## How to find the common difference of an arithmetic sequence?

By definition, the common difference of an arithmetic sequence is the fixed number you add or subtract to find the following number of the sequence. Usually, we use the first and second terms of the arithmetic progression to find the common difference. Thus,

d= $a_2-a_1$

where,

d is the common difference of the sequence;

$a_1$ is the first term of the sequence; and

$a_2$ is the first term of the sequence.

Generally, we can define the common difference of the sequence as:

d= $a_m-a_{m-1}$

where,

d is the common difference of the sequence;

$a_m$ is the any term in the arithmetic sequence; and

$a_{m-1}$ is the term before $a_m$.

Example #1

What is the common difference of the arithmetic sequence of 6, 9, 12, 15, 18, 21?

Solution

Step 1: Simply use the formula of d= $a_2-a_1$. Thus,

d = 9 – 6

d = 3

Step 2: Check if it is true for all the succeeding terms. Thus,

1. – 9 = 3; 15 – 12 = 3; 18 – 15 = 3; 21 – 15 = 3

Since 3 is the difference of all the consecutive terms, we can now conclude that the common difference is 3.

Therefore, the common difference of the sequence 6, 9, 12, 15, 18, 21 is 3.

Example #2

Determine the common difference of the arithmetic sequence __, 4, 10, 16, __, 28, 34, 40.

Solution

Step 1: Since there are missing terms in the given arithmetic sequence, we will use the formula d= am-am-1, wherein we can choose any two consecutive terms in the given arithmetic progression. Thus,

d = 16 – 10

d = 6

Step 2: Check if it is true for all the consecutive terms. Thus,

10 – 4 = 6; 34 – 28 = 6; 40 – 34 = 6

Since the result of subtracting all two consecutive terms is 6, we can conclude that it is their common difference.

Therefore, the common difference of the arithmetic sequence is 6.

## How to find the nth of an arithmetic sequence?

Usually, the nth term of an arithmetic sequence is defined by the first term and the arithmetic progression’s common difference. Thus,

$a_n = a_1 + (n – 1) d$

where,

$a_n$ is the nth term of the arithmetic sequence;

$a_1$ is the first term of the arithmetic sequence;

n is the position of the nth term in the arithmetic sequence; and

d is the common difference of the arithmetic sequence.

Generally, we can denote it as:

$a_n = a_m + (n – m) d$

where,

$a_n$ is the nth term of the arithmetic sequence;

$a_m$ is any term in the arithmetic sequence;

n is the position of the nth term in the arithmetic sequence;

m is the position of the mth term in the arithmetic sequence; and

d is the common difference of the arithmetic sequence.

Example #1

What is the 7th term of the arithmetic sequence if the first term is 5 and the common difference is 13?

Solution

Step 1: List all the information given in the problem.

$a_n = a_7; a_1 = 5$; d = 13; n = 7

Step 2: To find the 7th term of the arithmetic progression, use the formula

$a_n = a_1 + (n – 1) d$. By substitution, we will have:

$a_n = a_1 + (n – 1) da_7 = 5 + (7 – 1) 5a_7 = 5 + (6) 5 a_7 = 5 + 30a_7 = 35$

Therefore, the seventh term of the arithmetic progression is 35.

Example #2

Determine the 12th term of the arithmetic sequence if the fifth term is 23 and the common difference is 7.

Solution

Step 1: List all the information given in the problem.

$a_n = a_{12}; a_5 = 23$; d = 7; n = 12; m = 5

Step 2: To find the 12th term of the arithmetic progression, use the formula

$a_n = a_m + (n – m) d$. By substitution, we will have:

$a_n = a_m + (n – m) da_{12} = 23 + (12 – 5) 7a_{12} = 23 + (7) 7a_{12} = 23 + 49a_{12} = 72$

Example #3

Find the common difference and the unknown terms of the arithmetic sequence shown below.

9, 13, __, 21, 25, 29, __, 37, 41, …

Solution

Step 1: Determine the common difference of the arithmetic sequence. Thus,

d = 13 – 9

d = 4

Step 2: The missing unknown terms are the 3rd and 7th terms. Since the 2nd and 6th term of the arithmetic sequence is shown, we can simply add the common difference to the preceding term of the unknown terms. Thus;

$a_3 = a_2 + d$

$a_3 = 13 + 4a_3 = 17$

$a_7 = a_6 + d$

$a_7 = 29 + 4a_7 = 33$

Therefore, the common difference of the arithmetic progression is 4. The third and seventh term is 17 and 33, respectively.

Example #4

Find the common difference and the 20th term of the arithmetic sequence described below.

-21, -15, -9, -3, 3, …

Solution

Step 1: Determine the common difference of the arithmetic progression. Thus,

d = -15 – (-21)

d = -15 + 21
d = 6

Step 2: List all the information necessary to solve the problem.

$a_n = a_{20}; a_1 = -21$; d = 6; n = 20

Step 3: To find the 7th term of the arithmetic progression, use the formula

$a_n = a_1 + (n – 1) d$. Thus, by substitution,

$a_n = a_1 + (n – 1) da_{20} = -21 + (20 – 1) 6a_{20} = -21 + (19) 6 a_{20} = -21 + 114a_{20} = 93$

Therefore, the common difference is 6 and the 20th term is 93.

## How to formulate the general formula for the nth term of an arithmetic sequence?

The formula of finding the nth term, $a_n = a_1 + (n – 1)d$, yields the general form of the arithmetic progression as long as the first term and the common difference is known.

Say, for example, the sequence -21, -15, -9, -3, 3, … have a common difference that is 6. Then, using the formula, the general form of the sequence -21, -15, -9, -3, 3, … will be:

$a_n = a_1 + (n – 1) da_n = -21 + (n – 1) 6a_n = -21 + (6n – 6)a_n = 6n – 27$

Therefore, the general formula in finding the nth term of the sequence is -21, -15, -9, -3, 3, … is

$a_n = 6n – 27.$

Let’s look at the other examples below.

## Sum of an Arithmetic Sequence

### What is the sum of the arithmetic sequence?

An arithmetic series is the sum of the terms of a finite arithmetic progression. The arithmetic series or sum of the arithmetic sequence is defined by the formula

$S_n= \frac{n(a_1 + a_n)}{2}$

where,

$S_n$ is the sum of the finite arithmetic progression;

$a_1$ is the first term of the finite arithmetic progression;

$a_n$ is the last term of the finite arithmetic progression; and

n is the number of terms in the finite arithmetic progression.

### How to derive the formula for the sum of the arithmetic sequence?

To derive the formula $S_n= \frac{n(a_1 + a_n)}{2}$ , begin by expressing the arithmetic sequence in two different ways such that:

$S_n= a_1 + (a_1 + d) + (a_1 + 2d)+… +[a_1 + (n – 2) d] + [a_1 + (n – 1) d]S_n=[a_n – (n – 1) d]+ [a_n – (n – 2) d] +…+[a_n- 2d+ a_n- d + a_n$

Then, by adding the two arithmetic sequence, all terms involving d will be cancelled out. Thus, having the equation:

$2S_n = n (a_1 + a_n)$

Dividing both sides of the equation by 2 will result to:

$S_n = \frac{n (a_1 + a_n)}{2}$

Alternatively, if substitute the value of $a_n$ to $a_n = a_1 + (n – 1) d$, we will have the form:

$S_n = \frac{n}{2}[2{a_1} + (n+1)d]$

### How to solve the sum of an arithmetic sequence?

To solve for the arithmetic series or the sum of an arithmetic sequence:

1. Determine the first term and common difference of the arithmetic sequence.
2. Use the formula $S_n = \frac{n (a_1 + a_n)}{2}$ in finding the sum.

Example #1

What is the sum of the finite arithmetic sequence 21, 29, 37, 45, 53, 61?

Solution

Step 1: Determine the common difference of the arithmetic progression. Thus,

d = 29 – 21

d = 8

Step 2: List all the necessary information.

n = 6, $a_1$ = 21, $a_6$ = 61, d = 8

Step 3: Use the formula $S_n = \frac{n (a_1 + a_n)}{2}$ to solve for the arithmetic series. By substitution, the sum is:

$S_n = \frac{n (a_1 + a_n)}{2} S_6= \frac{6(21 + 61)}{2} S_6 = \frac{6(82)}{2} S_6 = \frac{492}{2} S_6 = 246$

Therefore, the sum of the finite arithmetic progression is 246.

Example #2

What is the sum of the first 50 counting numbers?

Solution

Step 1: Determine the common difference of the arithmetic sequence. Since the problem states that the we are looking for the sum of the first consecutive counting numbers, thus, d = 1.

Step 2:  List all the necessary information.

n = 50, $a_1$ = 1, $a_{50}$ = 50, d = 1

Step 3: Use the formula $S_n = \frac{n (a_1 + a_n)}{2}$ to solve for the arithmetic series. By substitution, the sum is:

$S_n = \frac{n (a_1 + a_n)}{2} S_{50}= \frac{50(1 + 50)}{2} S_{50} = \frac{50(51)}{2} S_{50} = \frac{2550}{2} S_{50} = 1275$

Therefore, the sum of the first 50 counting numbers is 1275.

Example #3

If we make an arithmetic progression that starts counting the year from 1990 until 2021, what will be the sum of this arithmetic progression?

Solution

Step 1:  Determine the common difference of the arithmetic sequence. Since the problem states that the we are looking for the sum of the arithmetic progression if we count by years, thus, d = 1.

Step 2: List all the necessary information.

n = 32, $a_1 = 1990, a_{32} = 2021$, d = 1

Step 3: Use the formula $S_n = \frac{n (a_1 + a_n)}{2}$ to solve for the arithmetic series. By substitution, the sum is:

$S_n = \frac{n (a_1 + a_n)}{2} S_{32}= \frac{32(1990+ 2021}{2} S_{32} = \frac{32(4011)}{2} S_{32} = \frac{128 532}{2} S_{32} = 64 176$

Therefore, the sum of the arithmetic progression is 64 176.

## What is the importance of arithmetic sequence?

Mathematicians were inspired by nature to attempt to explain these patterns in nature, to develop mathematical models, and gain a basic understanding of geometric shapes and structures.

In real life, the arithmetic sequence is essential because it enables us to understand concepts through patterns. An arithmetic sequence is an excellent starting point for describing various things, such as time, which has a common difference of one hour. An arithmetic sequence is also necessary for simulating recurring events.

Additionally, mathematical sequences and series are used in:

• stacking cups, chairs, bowls;
• pyramid-like patterns in which objects are constantly increasing or decreasing in size;
• business and financial analysis to aid in decision-making;
• determine the optimal solution to a given problem; and
• quantitative analysis is used by organizations to perform risk assessment and management, investment decision-making, and pricing, among other critical functions.