# Arithmetic Progression

In this section, we will learn about Arithmetic Progression.

We often come across the word “ sequence” in our everyday life. A sequence can be defined as a proper arrangement of something in a certain pattern or order.

Progression– can be defined as the sequence in which each term follows a particular pattern.

### Arithmetic Progression

Arithmetic Progression(A.P.) – In an arithmetic progression, the difference between the two consecutive terms is always the same.
In other words, It is a sequence in which each term increases or decreases by a fixed constant.

The fixed difference between the consecutive terms is called the “common difference” of the Arithmetic Progression.It is denoted by “d”.
It is to be noted, that this difference can be positive, negative, or zero.

Each number in the Arithmetic Progression is called a term.

In an Arithmetic Progression, if “a” is the first term and “d” is the common difference, Then
the general form of an arithmetic progression is given by,
a, a+d, a+2d, a+3d,…..

Also,If { a }_{ 1 }, { a }_{ 2 }, { a }_{ 3 },…. { a }_{ n } are in AP then common difference d is given by,

d= { a }_{ 2 }{ a }_{ 1 } = { a }_{ 3 }{ a }_{ 2 } = { a }_{ 4 }{ a }_{ 3 }=……= { a }_{ n }{ a }_{ n-1 }

When there are definite number of terms in an arithmetic progression then it is called finite arithmetic progression.
Moreover,  if there are indefinite number of terms in an arithmetic progression then it is called infinite arithmetic progression.

#### nth term of an Arithmetic Progression:

When in a finite arithmetic series-

“a” is the first term
“d” is the common difference
“l” is the last term
{ a }_{ n } is the nth term of the AP

Thus,

 nth term of the Arithmetic Progression is given by-                          { a }_{ n }  = a + (n-1) d

Also,

1. Three consecutive terms in an A.P is given by-
a – d, a , a+ d

2. Four consecutive terms in an A.P is given by-
a -3d, a-d, a+d, a+3d

#### Sum of n terms In an AP

The sum of first n term in an A.P. whose first term is “a” and common difference is “d” is given by-

#### { S }_{ n } = \frac { n } { 2 } [2a + (n-1)d]

Also , if “l” is the last term of A.P
then,

#### { S }_{ n } = \frac { n } { 2 } [a + l ]

⇒ The nth term of an AP is the difference of the sum to first n term and the sum to first (n-1) terms  of it.
i.e   { a }_{ n } ={ S }_{ n }{ S }_{ n-1 }

To summarise,

#### Arithmetic Mean:

Suppose there are two numbers a and b. We want to insert a number “A” between them such that a,A,b are in A.P. Here A is the arithmetic mean of the numbers a and b.
Also,
as a, A, b are in A.P.
⇒    A – a = b – A  {as the difference of consecutive numbers in AP are equal}

#### A = \frac { a+b }{ 2 }

Now, We can also insert n numbers { A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } between a and b. Such that a,{ A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } ,b are in A.P.

As a,{ A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } ,b are in AP.
Here the total number of terms is (n+2).
and b is the (n+2)th term.
So,
b = a +[(n+2)-1] d
⇒   b = a + (n + 1)d
Thus,

#### d=\frac { b-a }{ n+1 }

Thus common difference of this AP is d=\frac { b-a }{ n+1 }

and the terms are
{ A }_{ 1 } =a + d = a+ (\frac { b-a }{ n+1 })

{ A }_{ 2 } =a + 2d = a+ 2(\frac { b-a }{ n+1 })
.
.
.
{ A }_{ n } =a + nd = a+ n(\frac { b-a }{ n+1 })