Arithmetic Sequence:

An arithmetic sequence is defined as a group of numbers in which any two consecutive or successive terms have a common difference.

**Example: **

In the sequence of following numbers:

3, 7, 11, 15, 19, ....

the common difference between any two **consecutive numbers** is 4.

Therefore, the numbers 3, 7, 11, 15, 19 ...are said to be in Arithmetic Sequence

**Denotation of terms in an Arithmetic Sequence**

In an Arithmetic Sequence, the following denotations are generally followed:

**a** denotes the first term,

**d**, the common difference and

**n**, the position of any term or a particular term.

Then the **n **terms of an **Arithmetic Sequence **are written as

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

In the above sequence of **n** terms

a is 1^{st} term,

a + d is 2^{nd} term,

a + 2d is 3^{rd} term, and

a + (n – 1) ×d is the **last **term

**a + (n – 1) × d** is also called the last term or the **n ^{th} ** term or still the

We will discuss below the formulae for finding any particular term and sum of any number of terms in an arithmetic sequence.

The nth term of a sequence denotes the position of a term or a particular term in the sequence.

Any term that is to be found is referred to as

Let the n terms of an Arithmetic Sequence with a

Any term in an Arithmetic Sequence is called general term or nth term and it is denoted as

To find this

a is the first term,

d is the common difference and

n is the position of the term.

1. The first three terms of an Arithmetic Sequence are 3, 7, 11 ...What is the 50

In the A.S. 3, 7, 11, the common difference is 4.

To find the 50th term, apply the above formula for nth term

a

In this the first term a = 3, common difference d = 4 and n = 50.

Applying the above formula, the 50

3 + (50 – 1) × 4 = 3 + 196 = 199

2. In an arithmetic sequence with a common difference of 5, the 25

Represent 25

Next, from the n

we can write, a

What needs to be found is the first term, a.

Applying the nth term formula, we get

180 = a + (25 – 1) × 5,

180 = a + 120, so, a = 60.

Again, from the nth term formula, the 15

a

Let the n terms of an Arithmetic Sequence be

a, a + d, a + 2d, ............, a + (n – 1)× d

If S

S

Now, let ‘l’ denote the last term, then

Since l = a + (n – 1)d, therefore

S

Now, first term, a = 2,

Common difference, d = 3,

and the last term is

l = a + (n – 1) × d is

2 + (51 – 1) × 3 = 2 + 150 = 152

Therefore, S = 51/2 [2 + 152] = 3927

** Arithmetic progression. **

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