##### Arithmetic Progression

### Definition

Arithmetic Progression is the sequence of numbers such that the difference between the two successive terms is always constant. The difference is called the common difference. It is also known as Arithmetic Sequence.

The first term of the sequence is called the initial term which is denoted as ‘a’.

The difference by which the consecutive numbers of the sequence increases or decreases is called the Common Difference, which is labelled by ‘d‘.

The general term is the nth term of the arithmetic progression with the initial term ‘a’ and the common difference ‘d’ is as $ \displaystyle {{a}_{n}}=a+(n-1)d$.

**Example 1**: Find the nth term of 1,5,9,13,17,21,…….an if the number of terms are 13?

Solution: We have the arithmetic progression 1, 5, 9, 13, 17, 21,…….an and n = 13

By the general formula $ \displaystyle {{a}_{n}}=a+(n-1)d$

The first term is a = 1

Common difference d = 5 – 1 = 4

Therefore an = 1 + (13-1)×4

an = 1 + (12×4) = 49

### Finite and Infinite Arithmetic Progression

The arithmetic progression with a limited number of terms is called Finite Sequence. It has a last term.

Example: 1, 4, 7, 10, 13,………,40.

This is a finite sequence with a = 1 and d = 3.

The arithmetic progression with unlimited number of terms is called **Infinite Sequence**. It does not have a last term.

Example: 1,2,3,4,5,…………….

This is an infinite sequence with a=1 and d=1

### Positive and Negative Common Difference

Based upon the value of the common difference that the arithmetic progression is increasing or decreasing.

1. If the common difference ‘d’ is positive then the arithmetic progression will increase towards positive infinity and it will be an increasing sequence.

Example: 10,15,20,25,30,35,40,………..

This is an increasing sequence with a = 10 and d = 5

2. If the common difference ‘d’ is negative then the arithmetic progression will rise towards negative infinity and it will be a decreasing sequence.

Example: 10,5,0,-5,-10,-15,-20,-25,………….

This is a decreasing sequence with a = 10 and d = -5

### Properties of Arithmetic Progression

### 1. If the terms of the arithmetic progression are increased or decreased with the same number then the resultant sequence will also be an arithmetic progression.

Example: 1, 2, 3, 4, 5, 6, 7,…… is an arithmetic progression with a=1 and d=1.

If we add 3 to every term of the sequence then the new sequence will also be an arithmetic progression 4, 5, 6, 7, 8, 9, 10,…… with a = 4 and d = 1

2. If the terms of an arithmetic progression are multiplied or divided with the same number (non zero) then the resultant sequence will also be an arithmetic progression.

Example: 2, 4, 6, 8, 10, 12, 14,…….with a=2 and d=2.

If we multiply every term of the sequence with 4 then the new sequence will be also an arithmetic sequence.

8, 16, 24, 32, 40, 48, 56,…..with a = 8 and d = 8.

### Formula of the n^{th} term of an Arithmetic Progression

As we said above: The general term is the nth term of the arithmetic progression with the initial term ‘a’ and the common difference ‘d’ is as $ \displaystyle {{a}_{n}}=a+(n-1)d$

Example 2: Find the 20th term of the sequence 1, 3, 5, 7, 9, 11….?

Solution: We see that our sequence is an arithmetic progression because the difference of the terms is constant d=2, also our first term a=1.

The general formula of the nth term is $ \displaystyle {{a}_{n}}=a+(n-1)d$

In our case n=20 as we have to find the 20th term.

$ \displaystyle {{a}_{{20}}}=1+(20-1)2$

$\displaystyle ~~~=1+(19\times 2)=39$

So the 20th term is 39.

### Sum of an Arithmetic Progression

If ‘a’ is the first term and `d` is the common difference of the series with n number of terms, then the sum of the series will be $ \displaystyle {{S}_{n}}=\frac{n}{2}\left[ {2a+(n-1)\left. d \right]} \right.$

Example 3: Find the sum of the arithmetic progression 2, 5, 8, 11, 14, 17?

Solution: We have the first term a=2 and the common difference d=3 and n=6 the total number of terms of the series.

Now all we have to do is substitute the values in the formula

$ \displaystyle {{S}_{n}}=\frac{n}{2}\left[ {2a+(n-1)\left. d \right]} \right.$

$ \displaystyle =\frac{6}{2}\left[ {2(2)+(6-1)3} \right]$

$ \displaystyle ~=3\left[ {4+5\cdot 3} \right]$

$ \displaystyle ~~~~~~=3\left[ {4+15} \right]$

$\displaystyle ~~=3\cdot 19=57$

So the sum is 57.

If we know the last term an and the first term a1 of the series instead of the common difference ‘d’ then the sum of the series would be $ \displaystyle {{S}_{n}}=\frac{n}{2}({{a}_{1}}+{{a}_{n}})$

**Example 4: **Find the sum of the arithmetic progression 1, 5, 10, 15, 20 with first term a_{1}=1 and a_{n}=20 ?

**Solution: **We know the formula is $ \displaystyle {{S}_{n}}=\frac{n}{2}({{a}_{1}}+{{a}_{n}})$

Since we have the first term, the last term and the number of terms now all we have to do is substitute the values on the formula.

$ \displaystyle {{S}_{n}}=\frac{5}{2}(1+20)$

$ \displaystyle ~=\frac{5}{2}\cdot 21$

$ \displaystyle =\frac{{105}}{2}=52.5$

So the sum is 52.5