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# Arithmetic Sequence

Top
 Sub Topics 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.

## Definition

A sequence a$_{1}$, a$_{2}$, ........,a$_{n}$ is called an arithmetic sequence when a$_{2}$ - a$_{1}$ = a$_{3}$ - a$_{2}$ = ........ = a$_{n}$ - a$_{n-1}$ .

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.

## Recursive formula for Arithmetic Sequence

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

## Formula

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 [$\frac{n(n+1)}{2}$]$^{2}$

## Sum of Arithmetic Sequence

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]

## Example Problems

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