**Introduction**

An arithmetic progression is a sequence in which the next term in the sequence is obtained by adding a constant to each term. The constant added is called the common difference. It is a sequence such that the difference between any two consecutive terms in the sequence is always a constant.

Suppose, n1, n2, n3……..nn are the

terms of an arithmetic progression sequence.

Then, n2 = n1 + d, n3 = n2 + d and so on.

Where n1 = the first term and d is the common difference

**Arithmetic Progression Examples**

Verify if the following sequence 3, 6, 9, 12, 15 is an arithmetic progression or not.

For this sequence to be an arithmetic progression sequence, the common difference between the consecutive terms should be constant.

Common difference (d) = n2 – n1 must equal to n3 – n2 and so on.

In this sequence, d = 6 – 3 = 3, 9 – 6 = 3, 12 – 9 = 3, and 15 – 12 = 3.

The difference between consecutive terms is constant. Hence, the above sequence is an arithmetic progression.

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**Arithmetic Progression Formula**

To understand the arithmetic progression formula, one should be familiar with the terminologies used in the formula.

**First-term**

As the name states, the first term is the first term of the sequence, which is usually represented by n1. For example, in the 5, 12, 19, 26, 33 sequence, the first term is 5.

**Common Difference**

A common difference is the fixed number that is added or subtracted between two consecutive terms (except the first term) in the arithmetic progression. It is denoted by ‘d’.

For example, if n1 is the first term, then:

n2 = n1 + d

n3 = n2 + d and so on

**Arithmetic Progression Formula to Find the** **General Term or n****th**** Term**

The general term or nth term in an arithmetic progression is found by:

**N****n**** = a + (n-1) *d**

where ‘a’ is the first term and ‘d’ is a common difference.

So, 1st term, N1 = a + (1-1) *d

2nd term, N2 = a + (2-1) *d

3rd term, N3 = a + (3-1) *d

By computing ‘n’ terms in the above formula, we get the general form of an arithmetic progression.

a, a + d, a + 2d, a + 3d, …… a + (n-1) *d

**Arithmetic Progression Formula to Find the** **Sum**

The **arithmetic progression formula** for the sum of ‘n’ terms where ‘a’ is the first term and ‘d’ is a common difference is as follows.

When the nth term is unknown:

**S****n**** = (n/2) * [2a + (n − 1) * d]**

When the nth term is known:

**Sn = (n/2) * [a****1**** + a****n****]**

Formula derivation

Let us assume that ‘t’ is the nth term of the series and Sn is the sum of first n terms in an arithmetic progression: a, (a + d), (a + 2d), …., a + (n – 1) * d.

Then,

Sn = a1 + a2 + a3 + ….an-1 + an

Substituting the terms in the above formula, we get

Sn = a + (a + d) + (a + 2d) + …….. + (t – 2d) + (t – d) + t …(1)

After writing the equation (1) in the reverse order

Sn =t + (t – d) + (t – 2d) + …….. + (a + 2d) + (a + d) + a …(2)

Now, add equation (1) and (2), we get

2Sn = (a + t) + (a + t) + (a + t) + …….. + (a + t) + (a + t) + (a + t)

2Sn = n * (a + t)

Sn = (n/2) * (a + t) …(3)

Let us replace the last term ‘t’ by the nth term in equation 3, we get,

nth term = a + (n – 1) * d

Sn = (n/2) * {a + a + (n – 1) * d}

Sn = (n/2) * {2a + (n – 1) * d}

**Example**

If you are asked to find the sum of the first 30 terms of a sequence 5, 11, 17, 23, ……

Solution:

a = 5, d = a2 – a1 = 11 – 5 = 6

Sn = (n/2) * {2a + (n – 1) * d}

Sn = (30/2) * (2 * 5 + (35 – 1) * 6}

Sn = (15) * (10 + 204)

Sn = 15 * 214

Sn = 3210

**Conclusion**

In mathematics, an arithmetic progression is a series of numbers where the difference between two consecutive terms is always constant. We can find multiple examples of arithmetic progression in our daily life. For example, enrollment numbers of students in a batch, months in a year, etc.

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