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:
- $$3,7,1,15, \ldots $$
- $$8,2, – 4, – 10, \ldots $$
- $${a_1},{a_1} + d,{a_1} + 2d,{a_1} + 3d, \ldots $$
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
$${a_{20}} = 3 + \left( {20 – 1} \right)2 = 3 + 79 \times 2$$
$$ = 3 + 38 = 41$$