__Arithmetic Progression__
Let us
consider the following lists of numbers

1.) 1,2,3,4,5,6.....

2.) -1,-2,-3,-4.....

3.) 100,90,80,70....

4.)1,1,1,1........

5.)0.5,1, 1.5, 2 ,2.5...

Here each
number is called term.

In 1.) Each term is 1 more than the preceding term.

In 2.) Each term is 1 less than the preceding term.

In 3.) Each term is 10 less than preceding term.

In 4.) Each term is constant or can be obtained by adding or
substracting 0 to preceding term.

In 5.) Each term is 0.5 more than preceding term.

From above all, we can easily find that all of the terms are
obtained by adding or substracting

some constant number to the preceding term.Such series of
numbers is called Arithmetic Progression.

Definition:

**An Arithmetic Progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except first term.**

**This fixed number is called common difference and denoted by**

__d.__
General Form
of A.P.

Let first term be a and common difference be d then

A.P. will be

**a, a+d, a+2d, a+3d, a+4d and so on**

If no of terms are
known series is called finite
series.

If no of terms are not known then series is called infinite series.

E.g.

Finite
A.P. 2,5,8,11,14

Non-finite
A.P. 1,3,5,7,9......uncountable terms

In the last A.P.

3-1=2

5-3=2

7-5=2
and so on

It can be seen that d can be found by substracting a number
in A.P. from next number to it.

Let Kth no be a(k)

then

d=a(k+1)-a(k)

Look at the following series

1,1,3,3,5,7

Here difference between two consecutive terms is not same
therefore it is not A.P

Solved
Examples

Q! Which of
the following form A.P.

1.)4,8,12,16

2.)1,-1,-3,-5

3.)-1,1,-1,1,-1

4.) 2,2,2,5,5,5

Solution

Check using
d=a(k+1)-a(k)

1.) 8-4=4

12-8=4

16-8=4

As in each case
difference is same therefore it is in A.P.

2.) -1-(1)=-2

-3-(-1)=-3+1=-2

-5-(-3)=-5+3=-2

Since difference in each case is same therefore this series
is also in A.P.

3.) 1-(-1)=2

-1-(1)=-2

1-(-1)=2

Since difference in
each case is not same therefore series is not in A.P.

4.) 2-2=0

2-2=0

5-2=3

As difference in each case is not same therefore series is
not in A.P.

How to find
nth term of A.P.

Suppose a moneybox initially has 100 rs.One day a child
deposits 5 rs and he keeps on doing so
every day.

Hence on 1

^{st}day money in moneybox 100rs
2

^{nd}day money in moneybox 100+5=105
3

^{rd}day money in moneybox 105+5=110 and so on
For 3

^{rd}day money can be found as 100+(3-1)*5=100+2*5=110
(In order to find 3

^{rd}term we add 2 times of common difference to the first term)
If we take a as first term and d as common difference then
above series can be written in general form as

Nth term
a(n)=a+(n-1)d

Solved
examples

1.
Find 7

^{th}term of given A.P.
2, 5,8…….

Sol.
a(7)=a+6d

Here first term=2

And common difference 5-2=3

Therefore
7

^{th}term
a(7)=2+6*3=20

2.
Which term of series is -84

15, 12, 9, ……………

Sol.
We know a(n)=a+(n-1)d

Here a(n)= -84

Common difference d=(12-15)=
-3

According to formula

a(n)=a+(n-1)d

-84= 15+(n-1)*(-3)

Or
-84-15=(n-1)*(-3)

-99=(n-1)(-3)

n-1= 33 or n=34

3.
In a flower bed there are 23 roses in first row
21 in second row and 19 in third row and so on. There are five roses in the
last row.Find the no of rows.

Sol. As
between each term there is difference of (21-23)=-2 therefor given rows form an A.P.

Where a(n)=5 common difference
d= -2

Therefor According to known formula

a(n)=a+(n-1)d

5=23+(n-1)(-2) or
-18=(n-1)(-2)

Or n-1=9

n=10

Sum of n terms of an A.P.

Consider following A.P.

a, a+d, a+2d, a+3d, a+4d………..

General formula to find sum of such n terms
can be given by

S(n)=n{2a+(n-1)d}/2

It can also be written as

S(n)=n{a+a+(n-1)d}/2

Or S(n)=n{a+l}/2 where l is
last term of series given by l=a+(n-1)d

Based on it we can also find sum of n positive integers as

Series of n positive no.

1, 2, 3, 4, 5, 6, 7, 8…….

Then
S(n) =n{2*1+(n-1)*1}/2

Or
S(n)=n{2+n-1}/2

S(n)=n(n+1)/2

Solved examples

1.
Find sum
of 24 terms of the list of no where nth term is given by

a=3+2n

Sol. Here First term a(1)=3+2*1=5

Second
term a(2)=3+2*2=7

Third term a(3)=3+2*3=9 and so on

Hence 24

^{th}term will be the last term and given by
a(24)=3+2*24=51

Using formula

S(n)=n{a+l}/2

S(24)=24(5+51)/2 =12*56

S(24)=672