# 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
${a_{20}} = 3 + \left( {20 – 1} \right)2 = 3 + 79 \times 2$
$= 3 + 38 = 41$