Arithmetric Progression

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....
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.

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.

Finite A.P.          2,5,8,11,14
Non-finite A.P.      1,3,5,7,9......uncountable terms

In the last A.P.
               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)
Look at the following series
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.
4.) 2,2,2,5,5,5
Check using d=a(k+1)-a(k)
1.)   8-4=4
 As in each case difference is same therefore it is in A.P.
2.)  -1-(1)=-2
Since difference in each case is same therefore this series is also in A.P.
3.)  1-(-1)=2
 Since difference in each case is not same therefore series is not in A.P.
4.)  2-2=0
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 1st day money in moneybox   100rs
2nd day money in moneybox   100+5=105
3rd day money in moneybox    105+5=110 and so on
For 3rd day money can be found as     100+(3-1)*5=100+2*5=110
(In order to find 3rd 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 7th term of given A.P.
2, 5,8…….

Sol.  a(7)=a+6d
       Here first term=2
And common difference  5-2=3
Therefore  7th term

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

-84= 15+(n-1)*(-3)
Or     -84-15=(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

5=23+(n-1)(-2)         or      -18=(n-1)(-2)
   Or        n-1=9

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

It can also be written as


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

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

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 24th term will be the last term and given by

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