An Introduction to Arithmetic Progression

A sequence of numbers following a special pattern is usually known as a progression. The arithmetic progression is one of the examples of this kind of sequence. In this section, we will learn about arithmetic progression.

What is an Arithmetic Progression

A sequence of numbers is called an arithmetic progression if the differences between the two consecutive terms are the same. For example,  consider the sequence \[1, 8, 15, 22, 29, \cdots.\] Note that the differences of each two consecutive terms are the same which is $7.$ So this sequence is an example of an arithmetic progression.

Remark: The arithmetic progression is abbreviated as AP.

 

General Form of an Arithmetic Progression

Let $a$ denote the first term of an arithmetic progression. If the difference between consecutive terms is $d$, then we have:

$2$nd term: $a+d$

$3$rd term: $a+d+d$ $=a+2d$

$4$th term: $a+2d+d$ $=a+3d$

and so on. Thus the general form of an arithmetic progression is \[a, a+d, a+2d,a+3d, \cdots .\] In the above general form of an AP, $a$ is called the first term and $d$ is called the common difference.

 

Terms and Notations

For a general arithmetic progression $a_1, a_2,$ $ a_3, \cdots$, the following notations are used to denote a few useful terms related to this AP:

•  $a_1$: first term

•  $d$: common difference. So we have $d$ $=a_2-a_1$ $=a_3-a_2=\cdots$

•  $a_n$: n-th term

•  $S_n$: the sum of the first n terms. So we have $S_n$ $=a_1+$ $a_2+$ $\cdots +a_n$

 

Examples of Arithmetic Progressions

(i) Lets consider the sequence $7, 7,$ $7, 7, \cdots.$ Here the first term is $a_1=7$ and the difference between the consecutive terms is $d=0$ as $7-7=0.$ So it is an AP with the first term $7$ and with the common difference $0.$ This type of AP is known as a constant AP.

(ii) Next, consider $15, 10,$ $5, 0,$ $-5, \cdots.$ Note that the first term is $a_1=15$ and the common difference is $d=-5$ as $15-10$ $=10-5$ $=0-5$ $\cdots=-5.$ So it isan example of an AP with the negative common difference.

(iii) Similarly, the sequence $0, 5, 10,$ $15, \cdots$ is an AP with $a_1=0$ and $d=5.$

From the above three examples, we can conclude that the common difference of an arithmetic progression can be zero, negative and positive. ♣

 

n-th term of an Arithmetic Progression

Let us consider a general arithmetic progression $a_1, a_2,$ $ a_3, \cdots.$ If $d$ is the common difference of this AP, then we have the following relation:

$d=a_2-a_1$ $=a_3-a_2=$ $\cdots=a_n-a_{n-1}$

So we can deduce that

The first term is $a_1$ $=a_1+(1-1)d$

The second term $a_2$ $=a_1+d$ $=a_1+(2-1)d$

The third term $a_3$ $=a_2+d$ $=(a_1+d)+d$ $=a_1+(3-1)d$

The fourth term term $a_4$ $=a_3+d$ $=(a_1+2d)+d$ $=a_1+(4-1)d$

The fifth term term $a_5$ $=a_4+d$ $=(a_1+3d)+d$ $=a_1+(5-1)d$

$\quad \quad \vdots \quad \vdots$

The n-th term $a_n=a_1+(n-1)d$

Hence, the n-th term of an AP with the first term $a_1$ and with the common difference $d$ is given as follows: $a_n=$ $a_1+(n-1)d.$ In other words,

the n-th term of an AP = first term $+ (n-1) \times$ common difference.

 

Sum of the terms of an Arithmetic Progression

Let an AP has the first term $a$ and the common difference $d.$ Then the sum of the first n terms of the AP is given below:

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

Proof: Note that the AP has the form $a,$ $a+d,$ $a+2d, \cdots .$ Its n-th term $a_n=$ $a+(n-1)d.$ As $S_n$ denotes the sum of the first $n$ terms, we have:

$S_n=a +$ $(a+d)+$ $(a+2d) +$ $\cdots + [a+(n-2)d]$ $+[a+(n-1)d]$ $\cdots$ (I)

Writing the above sum in the reverse order, we obtain that

$S_n=[a+(n-1)d] +$ $[a+(n-2)d]+$ $\cdots + (a+d)$ $+a$ $\cdots$ (II)

(I) + (II) termwise, we deduce that

$2S_n=[2a+(n-1)d]+\cdots +$ $[2a+(n-1)d]$ (n terms)

$\quad \,\, =n[2a+(n-1)d]$

∴ $S_n=\frac{n}{2}[2a+(n-1)d]$

 

Remark: For an AP with the first term $a$ and the common difference $d$, the sum of the first n terms is given as follows:

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

$\quad=\frac{n}{2}[a+a+(n-1)d]$

$\quad=\frac{n}{2}[a+\{a+(n-1)d\}]$

$\quad=\frac{n}{2}[\text{first term}+\text{n-th term}]$

∴ If the n-th term is the last term of the AP, then the sum of the terms of that AP = n/2(first term + last term). ♣

 

Arithmetic Mean

Let $z$ be the arithmetic mean of $a$ and $b.$ Then $a, z$ and $b$ form an AP. So we must have that the common difference is equal to

$z-a=b-z$

⇒ $2z=a+b$

⇒ $z=\frac{a+b}{2}$

∴ the arithmetic mean of two numbers is half of the sum of the two numbers.

Remark: If $a, b$ and $c$ are in AP, then $c$ is the arithmetic mean of $a$ and $b.$ ♣

 

Formulas of an Arithmetic Progression

We will list all the arithmetic progression formulas (AP formulas) in one place.

Let $a$ denote the first term of an arithmetic progression (AP) with common difference $d.$ Then we have the following:

(i) The AP has the form $\{a, a+d, a+2d, \cdots\}$

(ii) n-th term $=a+(n-1)d$

(iii) Sum of the first n terms $=\frac{n}{2}[2a+(n-1)d].$

(iv) If the last term of the AP is $l$, then the sum of the terms of the AP $= \frac{n}{2}(a+l).$

If three terms form an AP, then we should assume the numbers as

$a-d,$ $a$ and $a+d$

If four numbers form an AP, then one can assume the numbers as

$a-3d,$ $a-d,$ $a+d$ and $a+3d$

The sum of the first n natural numbers $=\frac{n(n+1)}{2}$

The sum of the squares of the first n natural numbers $=\frac{n(n+1)(2n+1)}{6}$

The sum of the cubes of the first n natural numbers $=\{\frac{n(n+1)}{2}\}^2$

 

Solved Problems of Arithmetic Progression
Problem 1: If an AP has the first term $4$ and the common difference $5$, then write down few terms of the AP.

Solution:

First term $=4$, Second term $=4+5$ $=9$

Third term $=9+5$ $=14$, Fourth term $=14+5$ $=19$

Fifth term $=19+5$ $=24$, Sixth term $=24+5$ $=29$

and so on. So the AP is $\{4, 9, 14, 19, 24, 29, \cdots\}$ ♣

 

Problem 2: Find the tenth term of the AP $\{12, 6, 0, \cdots\}$

Solution:

Here the first term $a=12$ and the common difference $d=$ $6-12$ $=-6$

So the tenth term is

$a_{10}$$=a+(10-1)d$ $=a+9d$ $=12+9 \times (-6)$ $=12-54$ $=-42$ ♣

 

Problem 3: If the sum of second and tenth terms of an arithmetic progression is equal to $12,$ find the sum of fourth, sixth, and eighth terms.

Solution:

Let $a$ and $d$ denote the first term and the common difference of the given arithmetic progression. We know that the n-th term of the sequence is $t_n=a+(n-1)d.$ Given that $$t_2+t_{10}=12.$$
This implies that $a+d+a+9d=12$
$\Rightarrow 2a+10d=12$
$\Rightarrow 2(a+5d)=12$
$\Rightarrow a+5d=6 \cdots (i)$
Now the sum of the fourth, sixth and eighth term is
$=t_4+t_6+t_8$
$=a+3d+a+5d+a+7d$
$=3a+15d$
$=3(a+5d)$
$=3 \times 6$ by (i)
$=18$
So the desired sum will be $18.$

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