ARITHMETICAL PROGRESSION (AP)

Consider the following list of numbers – 

(1)1,2,3,4,….. 

(2)200,150,100,50,… 

(3)-5,-5,-5,……. 

Each of the numbers in the list is called term . 

In (1) , each term is 1 more than the term preceding it . 

In (2) , each term is 50 less than the term preceding it . 

In (3) , all the terms in the list is -5 , that is , each term is obtained by adding (or subtracting ) 0 to the term preceding it . 

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 the first term. 

This fixed number is called the common difference of the AP. 

Also , common difference can be positive , negative or zero . 

Finite and infinite AP— 

An AP which contains finite number of terms is called a finite AP. 

An AP which is not finite is called an infinite AP. 

General Term of an AP ( nth term of an AP)— 

Let us denote the first term of an AP by a1 , second term a2, ……., nth term by a_n and the common difference by d . 

Then the AP becomes  a1 , a2 , ……., a_n 

So ,

         a2 – a1 = a3 – a2 = …….= a_n – a_n-1 = d  

If a is the first term and d is the common difference of an AP , then it can also be represented as  

              a , a+ d , a+2d , a+3d, ………. 

This is called the general form of an AP. 

Thus , the nth term an of the AP with first term a and common difference d is given by  

                           a_n = a + (n -1) d 

an is also called the general term of the AP. 

If l represents the last term of an AP , then  

               l = a + (n- 1) d  

Note – 

r^th term from the first term = a + (r- 1)d     ……..(1) 

r^th term from the last term = l – (r- 1)d …..(2) 

Adding (1) and (2) = a + l 

Solved Examples -

(1)Find the 56^th term of an AP   3 , 9,15,21,…..? 

Sol.    Here a = 3 , d = 6 , n = 56 ,  

Therefore , 56^th term = a+ (n -1)d 

                                       = 3 + (56-1)×6 

                                        = 3 + 55×6   

                                        = 3 + 330 = 333 

(2)Find the value of n in the AP  3, 7, 11,15,…..,499 ?

Sol.    Here a = 3 and d= 7-3 = 4 

Now,           l = a +( n-1)d 

                     499 = 3+(n-1)×4 

                     499 = 3 + 4n -4 

                       4n = 500 

                         n = 125 

(3) In an AP , if 8^th term is 15 and 25^th term is 49 , then find the value of first term ?

Sol.     Let the first term is a and common difference is d . 

Then ,               15= a + (8-1)d = a + 7d       ……(1) 

  Also ,                49 = a +(25-1)d = a +24d  …….(2) 

Subtracting (1) from (2) , we have   17d = 34  

                                                           Or   d = 2 

Putting d = 2 in (1) we have ,  a = 1 

Thus , first term of an AP is 1 . 

(4) Find whether following series is in AP or not ?

(a)2,5,8,11,… 

(b)1,4,6,7,…… 

Sol.(a)         Here ,   5-2 = 3 

                                   8-5= 3 , 11-8= 3  

From the above , we see that the given series is in AP. 

       (b).         Here ,      4-1= 3 

                                 6-4= 2 

                                 7-6= 1 

Given series is not in AP. 

(5)How many terms are in the AP   4,7,10,13,………,148. 

Sol.     Here , a= 4 , d = 7-4 = 3 and l = 148 

Thus,            l = a + (n-1)d 

                     148 = 4 + (n-1)×3 

                Or   148 = 4 + 3n – 3 

               Or     148 = 3n + 1 

                    Or     3n = 147 

                    Or      n = 49 

Therefore , number of terms in the AP is 49 . 

Sum of first n terms of an AP -

Consider the AP  

             a , a+ d , a+2d , ………. 

Where a is first term and d is the common difference . 

The n^th term of this AP is a + (n-1)d . 

Let S denote the sum of the first n terms of the AP , then  

            S = a +( a+ d) + (a +2d) +…….+(a + (n-1)d). 

So , the sum of the first n terms of an AP is given by   

            S = n/2 [ 2a + (n-1)d ] 

Remark – 

If there are only n terms in an AP and l is the last term , then  

      S = n(a +l ) /2 

Note – 

(1)If a , b , c are in AP , then b =( a+ c )/2 and b is called the arithmetic mean of a and c . 

(2)If a, a1 ,a2 , a3 , …….., an , b are in AP , then we say that a1 ,a2 , a3 , ……,a_n are the arithmetic means between a and b . 

(3)If we have to take  

        (a) three terms in an AP , we should take  

             (a -d) , a , (a +d) 

        (b)four terms of an AP , we should take  

             ( a-3d) , (a – d ) , (a +d) , (a +3d) 

        (c)five terms of an AP , we should take  

             (a-2d ) , (a -d) , a , (a+ d) , ( a +2d) 

Solved Examples – 

(1)Find the sum of 22 terms of an AP   6 , 9 , 12 , 15 , ……… ?

Sol.     

        Here , a = 6 , d = 3 , n = 22 , then  

                   S = 22 /2[ 2×6 + (22-1)×3]  

                       = 11 [12 + 21×3] 

                       = 11 [ 12 + 63] 

                        = 11×75 = 825 

(2)In an AP , the sum of 4^th and 10^th term is 34.Then , find the sum of 13 terms of an AP? 

Sol.    

Let a be the first term and d is the common difference of given AP. 

Then according to question , 4^th term + 10^th term = 34 

                                                   (  a + 3d) + (a +9d) = 34 

                                                       2a + 12d = 34 

Now , the sum of 13 terms of an AP = 13/2 [ 2a +(13-1)d] 

                                                                 = 13/2 [2a + 12d ] 

                                                                  = 13/2 × 34 = 221 

(3)If in an AP , the sum of 16 terms is 680 and common difference is 5 .Then ,find the first term? 

Sol.  

Here , sum of 16 terms = 680 and d = 5 

Let a be the first term . 

Then according to the question,     

            680 = 16/2 [ 2a + (16-1)×5 ] 

             680 = 8 [ 2a + 15×5 ] 

Or          680 = 8 [ 2a + 75 ] 

Or             2a = 10 

Or                 a = 5 

(4)Find 9 arithmetic means between 7 and 37 ? 

Sol.    

 Here     

 7 , a1 , a2 , ……., a9 , 37  is in AP where a1 , a2 , ….., a9 be the nine arithmetic means between 7 and 37 . 

Here , we see that in above AP , there are 11 terms , where a = 7 and 11^th term is 37,

that is , l = 37 . 

Let d be the common difference of an AP. 

Then ,      l = a +(n-1)d  

                37 = 7 + (11-1)d 

                  37 = 7 + 10d  

Or                d = 3 

Therefore , arithmetic means are  

     (7+3), (7+6),(7+9),(7+12),(7+15),(7+18),(7+21),(7+24),(7+27) 

Or   10 , 13,16,19,22,25,28,31,34 .