To recall, progression is a sequence of numbers in a particular order. If we observe in our regular lives, we come across progression quite often. Roll numbers of a class, days in a week or months in a year. Did you notice that counting numbers, even numbers or odd numbers, all follow a particular progression!

In mathematics, there are three different types of progressions. They are:

- Arithmetic Progression
- Geometric Progression
- Harmonic Progression

Here, let us have a detailed look of maths arithmetic progression along with formulas and examples.

## What is Arithmetic Progression in Maths?

Arithmetic progressions is a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.

A progression is a special type of sequence for which it is possible to obtain a formula for the nth term. The arithmetic progression is the most commonly used sequence in maths with easy to understand formulas.

**Formal Definitions:**We will refer arithmetic progression with the abbreviation AP from now on.

**Definition 1**: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.

**Definition 2**: The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.

Now, let us consider the sequence, 1, 4, 7, 10, 13, 16,… is considered as an arithmetic sequence with common difference 3.

To understand the concepts of AP:

Consider an AP to be:

a_{1}, a_{2}, a_{3}, ……………., a_{n}

In general, the common difference “ d ” can be obtained as

a_{2} – a_{1} = a_{3} – a_{2} = ……. = a_{n} – a_{n – 1} = d,

Where, “d” is the common difference.

The progression can also be written in terms of common difference, as follows

a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d.

where “a” is the first term of the progression.

Position of Terms |
Representation of Terms |
Values of Term |
---|---|---|

1 | a_{1} |
a + d = a + (1-1) + d |

2 | a_{2} |
a + 2d = a + (2-1) + d |

3 | a_{3} |
a + 3d = a + (3-1) + d |

4 | a_{4} |
a + 4d = a + (4-1) + d |

. | . | . |

. | . | . |

. | . | . |

. | . | . |

n | a_{n} |
a + (n-1)d |

## Arithmetic Progression Formula

The formula to find the nth term and sum of n terms of an arithmetic progression are given as follows:

nth Term of an Arithmetic Progression

The formula for finding the n-th term of a sequence

**a _{n} = a + (n − 1) × d**

Where,

a = First term

d = Common difference

n = number of terms

an = nth term

The finite portion of an AP is known as finite AP and therefore the sum of finite AP is known as arithmetic series. The behaviour of the sequence depends on the term common difference.

- If the term “d” is positive, then the members terms will grow towards positive infinity
- If the term “d” is negative, then the member terms grow towards negative infinity

### Sum of First “n” Terms or Arithmetic Series

For any progression, the sum of n terms can be easily calculated. For an AP, the sum of the first n terms can be calculated if the first term and the total terms are known. This formula is explained below:

Consider an AP consisting “n” terms.

Sum of n terms is given as:

**S = n/2[2a + (n − 1) × d]**

**Proof:**

Consider an AP consisting “n” terms having the sequence a, a + d, a + 2d, ………….,a + (n – 1) × d

Sum of first n terms = a + (a + d) + (a + 2d) + ………. + [a + (n – 1) × d] ——————-(i)

Writing the terms in reverse order,

we have: S = [a + (n – 1) × d] + [a + (n – 2) × d] + [a + (n – 3) × d] + ……. (a) ———–(ii)

Adding both the equations term wise, we have

2S = [2a + (n – 1) × d] + [2a + (n – 1) × d] + [2a + (n – 1) × d] + …………. + [2a + (n – 1) ×d] (n-terms)

2S = n × [2a + (n – 1) × d]

S = n/2[2a + (n − 1) × d]

### Arithmetic Progression Examples

Here, the problems to find the nth terms and sum of the sequence are solved in detail. Go through once and solve the practice problems.

**Example 1:** Find the value of n. If a = 10, d = 5, an = 95.

**Solution:**

Given, a = 10, d = 5, a_{n} = 95

From the formula of general term, we have:

a_{n} = a + (n − 1) × d

95 = 10 + (n − 1) × 5

(n − 1) = 17

n = 18

**Example 2:** Find the 20th term for the given AP:

3, 5, 7, 9, ……

**Solution:**

Given, a = 3, d = 5 – 3 = 2, n = 20

a_{n} = a + (n − 1) × d

a_{n} = 3 + (20 − 1) × 2

a_{n} = 3 + 38

⇒a_{n} = 41

**Example 3:** Find the sum of first 30 multiples of 4.

Solution: Given,

a = 4, n = 30, d = 4

We know,

S = n/2 [2a + (n − 1) × d]

S = 30/2[2 (4) + (30 − 1) × 4]

S = 15[8 + 116]

S = 1860

## Practice Questions

Question 1: Find the a_n and 10th term of the progression: 3,10,17,24.

Question 2: If a = 2, d = 3 and n = 90. Find an and Sn.

Question 3: The 7th term and 10th terms of an AP is 12 and 25. Find the 12th term.

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