Arithmetic Progression
Arithmetic progression is a particular type of sequence in which each term except the first term progress in a definite manner. It is denoted as A.P (arithmeticprogression).Examples :
(i) 2,5,8,11,14,...
(ii) 10,5,0,5,10, ...
(iii) 9,7,5,3,2,1,3,...
In each of the above sequence every term except the first term obtained by adding a fixed number(positive or negative) to the preceding number.
Example : In the first example (i), each term is obtained by adding 3 to preceding except the 1st term 2.
1st term = 2
2nd term = 2 + 3 = 5
3rd term = 5 + 3 = 8
4th term = 8 + 3 = 11
5th term = 11 + 3 = 14 and so on.
In the second (ii) example, subtract 5 from each preceding term.
1st term = 10
2nd term = 10  5 = 5
3rd term = 5  5 = 0
4th term = 0 5 = 5
5th term = 5  5 = 10 and so on.
In general A.P. can be defined as $a_{1},a_{2},a_{3}, ..., a_{n}$ is an A.P if there exists a constant number 'd' such that
1st term = $a_{1}$
2nd term = $a_{1} + d = a_{2}$
3rd term = $a_{2} + d = a_{3}$
4th term = $a_{3} + d = a_{4}$



nth term = $a_{n1} + d = a_{n}$ and so on.
The constant term 'd' is called the common difference of the A.P.
So A.P sequence is a, a+ d, a + 2d , a+ 3d,
Examples on Arithmetic Progression
1) Show that the sequence defined $a_{n}$ = 4n + 5 is an A.P. Also find the sequence.Solution : We have,
$a_{n}$ = 4n + 5
Put n = n + 1, we get,
$a_{n+1}$ = 4(n + 1) + 5
$a_{n + 1}$ = 4n + 4 + 5 = 4n + 9
Now, $a_{n + 1}  a_{n}$ = 4n + 9  (4n + 5)
= 4n + 9  4n  5
= 4
which is a constant number.
$a_{n + 1}  a_{n}$ is independent of 'n'
So the given sequence is in A.P.
As we know that ,$ a_{n}$ = 4n +5
∴ n= 1 ⇒ $a_{1}$ = 4(1) + 5 = 4 + 5 = 9
n = 2 ⇒ $a_{2}$ = 4(2) + 5 = 8 + 5 = 13
n = 3 ⇒ $a_{3}$ = 4(3) + 5 = 12 + 5 = 17 and so on.
So the sequence is 9,13,17,...
2) Show that the sequence defined by $a_{n} = 2n^{2}$ + 1 is not an A.P.
Solution : We have,
$a_{n} = 2n^{2}$ + 1
Put n = n +1
$a_{n +1} = 2 (n +1)^{2}$ + 1
= 2($n^{2}$ + 2n + 1) + 1
= 2 $n^{2}$ + 4n + 3
$a_{n + 1}  a_{n}$ = 2$n^{2}$ + 4n + 3  ( $2n^{2}$ + 1)
= 2$n^{2}$ + 4n + 2  $2n^{2}$  1
∴ $a_{n + 1}$  $a_{n}$ = 4n + 2
From the above it is clear that $a_{n + 1}$  $a_{n}$ is not independent of n, so the given sequence is not in A.P.
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