Arithmetic Sequence or Arithmetic Progression

An arithmetic sequence or progression (abbreviated as A.P) is a sequence in which each term after the first is obtained by adding a fixed number to the preceding term, which is called the common difference.

In other words, quantities are said to be in arithmetic sequence when they increase or decrease by a common difference. Thus each of the following series forms an arithmetic progression:

  1. $$3,7,1,15, \ldots $$
  2. $$8,2, – 4, – 10, \ldots $$
  3. $${a_1},{a_1} + d,{a_1} + 2d,{a_1} + 3d, \ldots $$

 

The common difference is found by subtracting any term of the series from the term which follows it. In example 1 above, the common difference is $$4$$; in the second it is $$ – 6$$; in the third it is $$d$$.

But $$1,2,4,8,16, \ldots $$ is not an A.P. Here the second term minus the first term is $$1$$, while the third term minus the second is $$2$$. The difference that is obtained does not remain the same.

 

The nth term of an arithmetic progression

Let $${a_1}$$ be the first term and $$d$$ be the constant difference. Then the second term is $${a_1} + d$$, and the third term is $${a_1} + 2d$$. In each of these terms, the coefficient of $$d$$ is $$1$$ less than the number of terms. Similarly, the 10th term is $${a_1} + 9d$$. The nth term is the $$\left( {n – 1} \right)th$$ term after the first term and is obtained after $$d$$ has added $$\left( {n – 1} \right)$$ times in succession. Hence, if $${a_n}$$ represents the $$nth$$ term, then
\[{a_n} = {a_1} + \left( {n – 1} \right)d\]

 

Example:

Find the seventh term of an A.P in which the first term is $$11$$ and the common difference is $$4$$.

Solution:
The seventh term may be designed as $${a_7}$$, we use
$${a_n} = {a_1} + \left( {n – 1} \right)d$$

as the formula and substitute for the variables to find$${a_7}$$.

Here $$n = 7$$,   $${a_1} = 11$$,   $$d = 4$$
$$\therefore $$   $${a_n} = 11 + \left( {7 – 1} \right)4 = 11 + 6 \times 4 = 35$$

Thus, the required seventh term is $$35$$.

 

Example:

Find the $$13th$$ term of the following arithmetic progression: $$7,17,27, \ldots $$

Solution:
Here $$n = 13$$,   $${a_1} = 7$$,   $$d = 17 – 7 = 10$$
$$\therefore $$ $${a_n} = {a_1} + \left( {n – 1} \right)d$$

This gives   $${a_{13}} = 7 + \left( {13 – 1} \right)\left( {10} \right) = 7 + 120 = 127$$

Thus, the required $$13th$$ term is $$127$$.

 

Example:
Find the $$20th$$ term of an A.P. whose $$3rd$$ term is $$7$$ and the $$8th$$ term is $$17$$.

Solution:
Using   $${a_n} = {a_1} + \left( {n – 1} \right)d$$, we have
$${a_3} = {a_1} + \left( {3 – 1} \right)d$$            $$ \Rightarrow 7 = {a_1} + 2d$$   ——- (1)

$${a_8} = {a_1} + \left( {8 – 1} \right)d$$            $$ \Rightarrow 17 = {a_1} + 7d$$ ——- (2)

Subtracting (1) and (2), we get
$$10 = 5d$$           $$ \Rightarrow d = 2$$

Putting the value $$d = 2$$ in (1), we obtain
$$7 = {a_1} + 2\left( 2 \right)$$          $$ \Rightarrow 7 = {a_1} + 4$$  $$ \Rightarrow {a_1} = 3$$

Putting $${a_1} = 3$$, $$d = 2$$, $$n = 20$$ in $${a_n} = {a_1} + \left( {n – 1} \right)d$$ we get
$${a_{20}} = 3 + \left( {20 – 1} \right)2 = 3 + 79 \times 2$$
$$ = 3 + 38 = 41$$