 # How to Find Arithmetic Progression

A progression is a sequence of numbers in which each term and the successive term will have a constant relation. Arithmetic Progression is the most commonly used sequence in mathematics. In an arithmetic progression, the difference between two successive terms will always be constant. For example, 2, 4, 6,……, 16. Here, the difference between any two successive numbers is a constant. As far as the JEE exam is concerned, arithmetic progression is an important topic. In this article, we will discuss AP.

## What is Arithmetic Progression?

A sequence in which the difference between two successive terms is always a constant is known as an Arithmetic Progression. It is abbreviated as AP.

## How to Write General Form of an AP?

The general form is represented by a, a + d, a + 2d, a + 3d,….. Where a denotes the first term and d is the common difference.

## Important Formulas

The nth term of an AP = a + (n-1)d

Common difference, d = an+1 – an

Sum of n terms of AP = (n/2)[2a + (n – 1)d]

Sum of n terms when first and last terms are given = (n/2) [first term + last term]

Sum of first n natural numbers = n(n + 1)/2

Sum of the squares of first n natural numbers = n(n + 1)(n + 2)/6

Sum of the cubes of first n natural numbers = [n(n + 1)/2]2

If a, b, c are in A.P, then the middle term b is called the arithmetic mean. b = (a+c)/2

### Example

Insert 4 numbers between 12 and 72 such that the resulting sequence is an Arithmetic Progression.

Solution:

Given first term a = 12

Last term = 72

We have to insert 4 terms between 12 and 72. So, total number of terms in AP = 6

So a + (n – 1)d = 72

12 + (6 – 1)d = 72

12 + 5d = 72

5d = 72 – 12

5d = 60

d = 60/5 = 12

So a2 = 12 + 12 = 24

a3 = 24 + 12 = 36

a4 = 36 + 12 = 48

a5 = 48 + 12 = 60

Hence, the four numbers between 12 and 72 are 24, 36, 48, 60.

Example 2.

The fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it. Prove that the resulting sum is the square of an integer.

Solution:

Let a-3d, a-d, a+d, a+3d be the first four terms of A.P

Here common difference = 2d

Then product + (2d)2 = (a-3d)(a-d)(a+d)(a+3d) + (2d)4

= (a2-9d2)(a2-d2) + 16d4

= (a2-5d2)2

We have (a2-5d2) = a2-9d2 + 4d2

= (a-3d)(a+3d) + (2d)2

= a2-5d2

a and d are integers. Hence (a2-5d2)2 is an integer.