Arithmetic sequence is a sequence of numbers such that the distinction between two consecutive members of the sequence as the same common difference. If we have a sequence as x1, x2 ,x3,......., and the sequence follows the rule as xj - xi = d for all values of i and j, then the sequence is called the arithmetic sequence. This type of sequence where any term except the first term is obtained by adding a fixed number to the previous term. In arithmetic sequence we just add the same values each time. |

That means arithmetic sequence is a sequence in which each term is obtained by adding a constant 'd' to the preceding term. This constant d is called the common difference of the arithmetic progression. If three numbers a, b, c are in arithmetic sequence we say

b - a = c - b or a + c = 2b, b is called the arithmetic mean between a and c.

If the initial term of an arithmetic sequence is a$_{1}$ and the common difference of successive members is

*d*, then the

*n*th term of the sequence a$_{n}$is given by,

a$_{n}$ = a$_{1}$ + d(n - 1 )

a$_{n}$: nth term in the sequence

a$_{1}$: First term

d: Common difference

n: Term number

The above formula is the recursive formula for an arithmetic sequence.

**The formula is explained with the help of an example**.

Consider 2, 5, 8, 11, 14, ... be an arithmetic progression in which d = 3 is the common difference.

1st term = 2, 2nd term = 5, 3rd term = 8.

So 2nd term - 1st term= 5 - 2 = 3,

3rd term - 2nd term = 8 - 5 = 3

Here the difference between a term and the preceding term is same that is always constant. This constant is called common difference.

Now in general an arithmetic progression series can be written as

a, a + d, a + 2d, ....

Where a is the first term and d is the common difference.

Thus 1st term (t$_{1}$)= a = a + (1 - 1)d

2nd term (t$_{2}$) = a + d = a + (2 - 1) d

..............................................................

Nth term (t$_{n}$)= a + (n - 1) d, where n is the position number of the term.

Some of the commonly used formulas in arithmetic sequence are given below:

**1)**Sum of first n natural numbers

S = $\frac{n(n +1)}{2}$

**2)**Sum of first n odd numbers

S = n$^{2}$

**3)**Sum of the squares of first n natural numbers is

S = $\frac{n(n+1)(2n+1)}{6}$

**4)**Sum of the cubes of first n natural numbers is [$\frac{n(n+1)}{2}$]$^{2}$ Let S be the sum, a be the 1st term and l the last term of an arithmetic progression if the number of terms are n, then t$_{n}$ = l. Let d be the common difference of the arithmetic progression.

Now S = a + ( a + d) + (a + 2d) + ........ + ( l - 2d) + ( l - d) + l

Again S = l + ( l - d) + ( l - 2d) +....... + (a + 2d) + ( a + d) + a

On adding the above, we have

2S = ( a + l) + ( a + l) +.........+ (a + l)

= n (a + l)

S = $\frac{n(a+l)}{2}$

The above formula is used to determine the sum of n terms of an arithmetic sequence when the first term a and the last term is given.

The below formula is used when the first term a, common difference d and the number of terms of an arithmetic sequence are given. (Last term is not given)

S = $\frac{n}{2}$[2a + (n - 1)d]

**Some problems based on arithmetic sequence are as follows:**

**Example1:**If 5th and 12th terms of an arithmetic progression are 14 and 35 respectively. Find first five terms of the sequence.

**Solution:**

Let a be the first term and 'd' be the common difference.

T$_{5}$ = a + 4d = 14

T$_{12}$ = a + 11d = 35

On solving the above two equations

7d = 21

d = 3

Now a = 14 - (4 * 3) = 14 - 12 = 2

Hence the required sequence is 2, 5, 8, 11, 14.

**Example 2:**Divide 69 into 3 parts are in arithmetic sequence and are such that the product of the first two parts is 483.

**Solution:**

Given that the 3 parts are in arithmetic sequence

Let the three parts which are in A.P be a - d, a, a + d.

Thus a - d + a + a + d = 69

3a = 69

a= 23

So the three parts are 23 - d, 23, 23 + d

Since the product of first two parts is 483, therefore, we have

23 (23 - d) = 483

23 - d = $\frac{483}{23}$ = 21

d = 23 - 21 = 2

Hence the three parts which are in arithmetic sequence are

23 - 2 = 21, 23, 23 + 2 = 25

Finally the parts are 21, 23, 25.

**Example 3:**Which term of the arithmetic progression

$\frac{3}{\sqrt{7}}$, $\frac{4}{\sqrt{7}}$, $\frac{5}{\sqrt{7}}$, ........ is $\frac{17}{\sqrt{7}}$?

**Solution:**a = $\frac{3}{\sqrt{7}}$, d = $\frac{4}{\sqrt{7}}$ - $\frac{3}{\sqrt{7}}$ = $\frac{1}{\sqrt{7}}$, $t_{n}$ = $\frac{17}{\sqrt{7}}$,

We can write,

$\frac{17}{\sqrt{7}}$ = $\frac{3}{\sqrt{7}}$ + (n-1) * $\frac{1}{\sqrt{7}}$

$\Rightarrow$ 17 = 3 + (n - 1)

n = 17 - 2 = 15

Hence 15th term of the A.P. is $\frac{17}{\sqrt{7}}$.

**Example 4:**Insert 4 arithmetic means between 4 and 324.

4, -, -, -, -, 324

**Solution:**Here a = 4, d = ?, n = 2 + 4 = 6, t$_{n}$ = 324

Now t$_{n}$ = a + (n - 1)d

324 = 4 + (6-1)d

320 = 5d

d = $\frac{320}{5}$

d = 64

The first arithmetic progression = 4 + 64 = 68

Second arithmetic progression = 68 + 64 = 132

Third arithmetic progression = 132 + 64 = 196

Fourth arithmetic progression = 196 + 64 = 260

**Example 5:**Calculate the sixteen term for the given sequence 5, 8, 11, 14, 17,....

Solution : The given sequence has a difference of 3 between each number.

We see that the first term a = 5, common difference d = 3

Using the formula we can now find the sixteen term in the sequence

t$_{n}$ = a + (n - 1)d

= 5 + (n - 1)3

= 5 + 3n - 3

= 3n + 2

For sixteenth term, plug in n = 16

t$_{16}$ = 3 * 16 + 2

t$_{16}$ = 50