This presentation covered following topics : 1. Introduction 2. Arithmetic Progression (AP) 3. Sum of Series in AP 4. Arithmetic and Geometric Mean 5. Geometric Progression (GP) 6. Sum of Series in GP 7. Relation Between AM, GM and HM and is useful for B.Com and BBA students.

1. Semester :I
Mr. Tushar J Bhatt
1
Fundamental of Mathematics - I
Unit- 1: Arithmetic and Geometric Progression
1. Introduction
Table of contents
2. Arithmetic Progression (AP)
3. Sum of Series in AP
4. Arithmetic and Geometric Mean
5. Geometric Progression (GP)
6. Sum of Series in GP
7. Relation Between AM, GM and HM

2. Semester :I
Mr. Tushar J Bhatt
2
Arithmetic and Geometric Progression
1. Introduction
Definition : Sequence / Progression
1
2
3
T h e first term o f an y seq u ece is d en o ted b y T ,
T h e seco n d term is d en o ted b y T ,
T h e th ird term is d en o ted b y T ,
T h e n term is d en o ted b y T .
th
n
A set of numbers arranged in a definite order according to some rule is
called a sequence .
For example: (1) 4, 7, 10, 13, 16, …..
Note:

3. Semester :I
Mr. Tushar J Bhatt
3
Arithmetic and Geometric Progression
1. Introduction
Definition : Series
1 2 1
T h e su m o f first 'n ' term s o f a series is d en o ted b y S ,
T h erefo re S ...
n
n n n
T T T T
    
The algebraic sum of the terms of a sequence or a progression is called
a series.
For example: (1) 4 + 7 + 10 + 13 + 16+……..
Note:

4. Semester :I
Mr. Tushar J Bhatt
4
Arithmetic and Geometric Progression
1. Introduction
Definition : Arithmetic Progression(AP)
1
2
3
T h e firs t te rm o f a n A P is d e n o te d b y =
T h e c o m m o n d iffe re n c e o f a n A P is d e n o t e d b y =
1 T e rm = T = (1 - 1)
2 T e rm = T = ( 2 - 1)
3 T e rm = T (3 1)
...
T e rm = T = ( 1)
s t
n d
r d
th
n
a
d
a d
a d
a d
n a n d


  
 
A sequence in which each term is obtained by adding a constant
number to its preceding term is called an Arithmetic Progression (AP)
and the constant number is called the common difference.
For example: (1) 2,6,10,14,18, …….. Is an AP because in which the
common difference is 4.
Note:

5. Semester :I
Mr. Tushar J Bhatt
5
Arithmetic and Geometric Progression
Sum of ‘n’ terms of an AP (theory)
1
2
3
T h e firs t te rm o f a n A P is d e n o te d b y =
T h e c o m m o n d ife re n c e o f a n A P is d e n o te d b y =
1 T e rm = T =
2 T e rm = T =
3 T e rm = T 2
...
T e rm = T = ( 1)
N o w th e s u m o f firs t '
s t
n d
r d
th
n
a
d
a
a d
a d
n a n d

 
 
1 2 3
n ' te rm s o f a n A P is
...n n
S T T T T    
       ( ) ( 2 ) ... ( ( 1) ) ....(1)n
S a a d a d a n d         
N o w eq ....(1 ) can b e arran g ed in reverse o rd er w e g et
         ( ( 1) ) ( ( 2) ) ... ( 2 ) ( ) ....(2)n
S a n d a n d a d a d a            

6. Semester :I
Mr. Tushar J Bhatt
6
Arithmetic and Geometric Progression
Sum of ‘n’ terms of an AP
           
     
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
C o n sid er eq ....(1 ) + eq ....(2 ) w e g et,
2 ( 1) ( 2 ) ... ( 1)
2 2 ( 1) 2 ( 1) ... 2 ( 1)
n
n
n tim es
S a a n d a d a n d a n d n
S a n d a n d a n d
                        
          
 
 
_
2 2 ( 1)
2 ( 1) .
2
n
n
S n a n d
n
S a n d
    
   
       ( ) ( 2 ) ... ( ( 1) ) ....(1)n
S a a d a d a n d         
         ( ( 1) ) ( ( 2) ) ... ( 2 ) ( ) ....(2)n
S a n d a n d a d a d a            

7. Semester :I
Mr. Tushar J Bhatt
7
Arithmetic and Geometric Progression
1. Introduction
 
T h e s u m o f 'n ' te rm s o f a n A P is g iv e n b y
= 2 ( 1)
2
n
n
S a n d 
Note:
Ex-1: Find the 50th term of an AP : 37, 33, 29,
25,…Solution 1 :
5 0
5 0
5 0
5 0
H e re g iv e n A P is : 3 7 , 3 3 , 2 9 , 2 5 , ....
th e re fo re 3 7 , 4 , 5 0 .
( 1)
3 7 (5 0 1) ( 4 )
3 7 4 9 ( 4 )
3 7 1 9 6
1 5 9
n
a d n
T a n d
T
T
T
T
   
   
     
    
  
  

8. Semester :I
Mr. Tushar J Bhatt
8
Arithmetic and Geometric Progression
Ex-2 : The 4th term of an AP is 19 and its 12th term is 51,
then find its 21st term.
Solution 2 :
4
1 2
H e re g iv e n th a t T 3 1 9 ...................(1)
a n d T 1 1 5 1 ...............(2 )
C o s id e r (1 ) - (2 ) w e g e t,
3 1 6
1 1 5 1
_ _ _ _ _ _ _ _ _ _ _ _
8 3 2
4 p u t in e q u a tio n ......(1 ) w e g e t,
(1) 3 ( 4 ) 1 9
1 9 1 2
a d
a d
a d
a d
d
d
a
a
a
  
  
 

 
  
 
  
  
 7

9. Semester :I
Mr. Tushar J Bhatt
9
Arithmetic and Geometric Progression
Solution 2 :
2 1
2 1
N o w v a lu e s o f 7 a n d 4 in e q u a tio n
2 0 7 2 0 4 7 8 0 8 7
8 7 is th e re q u ire d 2 1 te rm o f g iv e n A P .
s t
a d
T a d
T
 
       
 
Ex-3 : The 6th term of an AP is 47 and its 10th term is 75,
then find its 30th term.
Solution 3 :
6
1 0
H e re g iv e n th a t T 5 4 7 ...................(1)
a n d T 9 7 5 ...............(2 )
C o s id e r (1 ) - (2 ) w e g e t,
5 4 7
9 7 5
_ _ _ _ _ _ _ _ _ _ _ _
4 2 8
a d
a d
a d
a d
d
  
  
 

 
  
7 p u t in eq u atio n ......(1 ) w e g et,
(1) 5(7 ) 4 7
4 7 3 5
1 2
d
a
a
a
 
  
  
 

10. Semester :I
Mr. Tushar J Bhatt
10
Arithmetic and Geometric Progression
Solution 3:
3 0
3 0
N o w v a lu e s o f 1 2 a n d 7 in e q u a tio n
2 9 1 2 2 9 7 1 2 2 0 3 2 1 5
2 1 5 is th e re q u ire d 3 0 te rm o f g iv e n A P .
th
a d
T a d
T
 
       
 
Ex-4 : Find the sum up to the required number of terms
of the followings:
(a) 100, 93, 86, 79, …….( up to 20 terms )
(b) 7, 19/2, 12, 29/2, 17, …….( up to 30 terms )
Solution 4 (a) :
 
H e re th e g iv e n A P is 1 0 0 , 9 3 , 8 6 , 7 9 , .. .
T h e re fo re 1 0 0 , 7 , 2 0
W e k n o w th a t 2 ( 1)
2
n
a d n
n
S a n d
   
  

11. Semester :I
Mr. Tushar J Bhatt
11
Arithmetic and Geometric Progression
Solution 4 (a) :
 
 
 
 
 
2 0
2 0
2 0
2 0
2 0
N o w 2 ( 1)
2
2 0
2 1 0 0 ( 2 0 1) ( 7 )
2
1 0 2 0 0 (1 9 7 )
1 0 2 0 0 1 3 3
1 0 6 7
6 7 0
n
n
S a n d
S
S
S
S
S
  
      
    
   
  
 
Solution 4 (b) :
 
1 9 2 9
H e re th e g iv e n A P is 7 , , 1 2 , , 1 7 , ...
2 2
5
T h e re fo re 7 , , 3 0
2
W e k n o w th a t 2 ( 1)
2
n
a d n
n
S a n d
  
  

12. Semester :I
Mr. Tushar J Bhatt
12
Arithmetic and Geometric Progression
Solution 4 (b) :
 
 
 
3 0
3 0
3 0
3 0
3 0
3 0
N o w 2 ( 1)
2
3 0 5
2 7 (3 0 1)
2 2
( 2 9 5 )
1 5 1 4
2
2 8 1 4 5
1 5
2
1 7 3
1 5
2
2 5 9 5
2
1 2 9 7 .5
n
n
S a n d
S
S
S
S
S
S
  
  
        
  
 
   
 
 

  
  
 
 

13. Semester :I
Mr. Tushar J Bhatt
13
Arithmetic and Geometric Progression
Ex-5: The 4th term of an AP is 22 and its 10th term is 52,
find the sum of its 40 terms.
Solution 5 :
4
1 0
H e re g iv e n th a t T 3 2 2 ...................(1)
a n d T 9 5 2 ...............(2 )
C o s id e r (1 ) - (2 ) w e g e t,
3 2 2
9 5 2
_ _ _ _ _ _ _ _ _ _ _ _
6 3 0
5 p u t in e q u a tio n .....(1 )
(1) 3 2 2
3 5 2 2
2 2 1 5
7
a d
a d
a d
a d
d
d
a d
a
a
a
  
  
 

 
  
 
  
   
  
 

14. Semester :I
Mr. Tushar J Bhatt
14
Arithmetic and Geometric Progression
Solution 5 :
 
 
 
 
 
4 0
4 0
4 0
4 0
4 0
H e n c e w e g e t 7 , 5 a n d 4 0 .
N o w w e k n o w th a t 2 ( 1)
2
4 0
2 7 ( 4 0 1) 5
2
2 0 1 4 3 9 5
2 0 1 4 1 9 5
2 0 2 0 9
4 1 8 0
n
a d n
n
S a n d
S
S
S
S
S
  
  
     
    
   
  
 

15. Semester :I
Mr. Tushar J Bhatt
15
Arithmetic and Geometric Progression
Ex-6: Find the 40th term and the sum of first 40 terms of
the sequence 1, 3, 5, 7, …….
Solution 6 :
4 0
4 0
H e re th e g iv e n s e q u e n c e 1 , 3 , 5 , 7 , .... . is a n A P
a n d g iv e n th a t 1, 2
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
T o F in d : T
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
T 3 9 1 3 9 2 1 7 8 7 9
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
a d
a d
 
        
 
 
4 0
4 0
_ _ _ _ _ _ _ _
T o fin d :
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
w e k n o w th a t 2 ( 1)
2
4 0
2 1 ( 4 0 1) 2
2
n
S
n
S a n d
S
  
     

16. Semester :I
Mr. Tushar J Bhatt
16
Arithmetic and Geometric Progression
Solution 6 :
 
 
 
 
4 0
4 0
4 0
4 0
4 0
4 0
2 1 ( 4 0 1) 2
2
2 0 2 3 9 2
2 0 2 7 8
2 0 8 0
1 6 0 0
S
S
S
S
S
     
    
   
  
 
Note :
1n n n
T S S 
 

17. Semester :I
Mr. Tushar J Bhatt
17
Arithmetic and Geometric Progression
Solution 7 :
2
2
1
2
1
2
1
2
1
H e re g iv e n th a t 2 3
N o w re p la c e 'n ' b y 'n -1 ' w e g e t,
2 ( 1) 3 ( 1)
2 ( 2 1) 3 3
2 4 2 3 3
2 1
n
n
n
n
n
S n n
S n n
S n n n
S n n n
S n n




 
   
     
     
   
Ex-7: The sum of ‘n’ terms of an AP is find its
20th term.
2
2 3 ,n n
 
1
2 2
1
2 2
2 2
N o w w e k n o w th a t
W h e re 2 3 a n d 2 1
2 3 2 1
2 3 2 1
4 1
n n n
n n
n
n
n
T S S
S n n S n n
T n n n n
T n n n n
T n


 
    
     
     
  
2 0
2 0
2 0
2 0
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
T o fin d :
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
4 1
4 ( 2 0 ) 1
8 0 1
8 1
n
T
T n
T
T
T
  
  
  
  Answer

18. Semester :I
Mr. Tushar J Bhatt
18
Arithmetic and Geometric Progression
Geometric Progression
Definition : Geometric Progression(GP)
2 3
2 3
2
If is th e firs t te rm a n d is th e c o m m o n ra tio o f a G P ,
it c a n b e e x p re s s e d a s , , , , ...
. . ...... c o m m o n ra tio
a r
a a r a r a r
a r a r a r
i e r
a a r a r
   
If in a sequence the ratio of any term to its preceding term is constant, it
is called a geometric progression. The constant ratio is called the
common ratio.
For example: (1) 5, 15, 45, 135, 405, …….. Is GP because in which
the common ratio is 3.
Note:

19. Semester :I
Mr. Tushar J Bhatt
19
Arithmetic and Geometric Progression
(a) nth term of a Geometric Progression
2 3
1 1
1
2 1
2
3 1 2
3
1
If is th e firs t te rm a n d is th e c o m m o n ra tio
o f a G P , it c a n b e e x p re s s e d a s , , , , ...
F irs t te rm =
S e c o n d te rm =
T h ird te rm =
.
.
.
.
te rm =
th n
n
a r
a a r a r a r
T a r a
T a r a r
T a r a r
n T a r




 
 
 


20. Semester :I
Mr. Tushar J Bhatt
20
Arithmetic and Geometric Progression
(b) Sum of first n terms of a Geometric Progression
2 3 1
2 3 1
If is th e f ir s t te r m a n d is th e c o m m o n r a tio o f a G P ,
it c a n b e e x p r e s s e d a s , , , , ..., th e n
... ......(1)
E q ....( 1 ) m u ltip ly in g b y ' ' b o th s id e s w e g e t,
(1)
n
n
n
n
a r
a a r a r a r a r
S a a r a r a r a r
r
r S r


     
   
               
2 3 1
2 3 4
2 2 3 1
2 2 3 3
...
... .....( 2 )
C o n s id e r e q ....( 1 ) e q ....( 2 ) w e g e t,
...
.....
n
n
n
n n
n n
n
n n
a a r r a r r a r r a r r
r S a r a r a r a r a r
S r S a a r a r a r a r a r a r a r
S r S a a r a r a r a r a r a r a r


        
      

                      
          

(1 ) (1 )
(1 )
, 1
(1 )
( 1)
, 1
( 1)
n
n n
n
n
n
n
n
n
S r S a a r
S r a r
a r
S if r
r
a r
S if r
r
  
   

  


  


21. Semester :I
Mr. Tushar J Bhatt
21
Arithmetic and Geometric Progression
Ex-1: Find the required term of the following GP
(a) 2, 6, 18, 54,……(9th term )
(b) 1024, -512, 256, -128,…… (10th term)
1
9 1
9
8
9
9
9
H e re g iv e n G P is : 2 , 6 , 1 8 , 5 4 , .....
6 1 8 5 4
2 , 3 ....
2 6 1 8
th e re fo re th e fo rm u la fo r n o f G P is g iv e n b y
2 (3 )
2 (3 )
2 6 5 6 1
1 3 1 2 2
th
n
n
a r
T a r
T
T
T
T


      

  
  
  
 
Solution 1 (a) :
Answer

22. Semester :I
Mr. Tushar J Bhatt
22
Arithmetic and Geometric Progression
1
1 0 1
1 0
9
1 0
1 0
H e re g iv e n G P is : 1 0 2 4 , -5 1 2 , 2 5 6 , -1 2 8 , .....
5 1 2 2 5 6 1 2 8 1
1 0 2 4 , ....
1 0 2 4 5 1 2 2 5 6 2
th e re fo re th e fo rm u la fo r n o f G P is g iv e n b y
1
1 0 2 4
2
1
1 0 2 4
2
1 0 2 4
th
n
n
a r
T a r
T
T
T


 
       


 
    
 
 
    
 
   
1 0
1
5 1 2
2T
 
 
 
  
Solution 1 (b) :

23. Semester :I
Mr. Tushar J Bhatt
23
Arithmetic and Geometric Progression
Ex-2: Find the sum up to the required term of the
following GP (a) 1, 2, 4, 8, 16,……(up to 12 terms )
(b) 8, 4, 2, 1,…… (up to 10 terms)
1 2
1 2
1 2
1 2
H e re g iv e n G P is : 1 , 2 , 4 , 8 , 1 6 , .....
2 4 8 1 6
1, 2 .... a n d 1 2
1 2 4 8
S o 2 1 th e n w e a p p ly th e fo rm u la
(r 1)
1
1 ( 2 1)
2 1
4 0 9 6 1
1
4 0 9 5
n
n
a r n
r
a
S
r
S
S
S
        
 



 
 


 
 
Solution 2 (a) :
Answer

24. Semester :I
Mr. Tushar J Bhatt
24
Arithmetic and Geometric Progression
1 0
1 0
H e re g iv e n G P is : 8 , 4 , 2 , 1 , .....
4 2 1
8 , .... a n d 1 0
8 4 2
1
S o 1 th e n w e a p p ly th e fo rm u la
2
(1 r )
1
1
8 1
2
1
1
2
n
n
a r n
r
a
S
r
S
      
 



  
      
 

Solution 2 (b) :

25. Semester :I
Mr. Tushar J Bhatt
25
Arithmetic and Geometric Progression
Solution 2 (b) :
1 0
1
8 1
1 0 2 4
1
2
S
 
  
 
 
1 0
1 0 2 3
8
1 0 2 4
1
2
S
 
  
 
 
1 0
1 0 2 3 2
8
1 0 2 4 1
S
 
    
 
1 0
1 0 2 3 2
1 2 8 1
S   1 0
1 0 2 3
6 4
S 

26. Semester :I
Mr. Tushar J Bhatt
26
Arithmetic and Geometric Progression
Ex-3: The 4th term of a GP is 4 and product of 2nd and
4th terms is 1. Find the 6th term and sum of first 6
terms.
3
4 3
2
2 4
3
2 4
2
4
2
H e re S u p p o s e th a t th e firs t te rm o f G P i s a n d
c o m m o n ra tio is
4
T h e re fo re 4 ......(1)
T h e S e c o n d te rm
G iv e n th a t 1
1
1
1
1
.......( 2 )
a
r
T a r a
r
T a r
T T
a r a r
a r
a
r
a
r
   

 
  
 
 
 
Solution 3 :

27. Semester :I
Mr. Tushar J Bhatt
27
Arithmetic and Geometric Progression
2 3
3 2
2 2
C o m p a re ....(1 ) & ....(2 ) w e h a v e
1 4
4
4
A n d
1 1 1
( 4 ) 1 6
r r
r r
r
a
r

 
 
  
Solution 3 :
6
1
N o w 4 a n d
1 6
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
T o fin d : T
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
r a 

28. Semester :I
Mr. Tushar J Bhatt
28
Arithmetic and Geometric Progression
Solution 3 :
 
5
6
5
6
6
6
T
1
T 4
1 6
1 0 2 4
T
1 6
T 6 4
a r 
 
   
 
 
  1st Answer
6
T o fin d :
N o w h e re 4 1 th e n w e a p p ly th e fo rm u la
(r 1) (r 1)
1 1
n n
n n
S
r
a
S S a
r r
 
 
   
 

29. Semester :I
Mr. Tushar J Bhatt
29
Arithmetic and Geometric Progression
Solution 3 :
2nd Answer
6
6
6
6
6
6
1 ( 4 1)
1 6 4 1
1 ( 4 1)
1 6 3
4 0 9 6 1
4 8
4 0 9 5
4 8
S
S
S
S
 
   
 
 
   
 

 
 

30. Semester :I
Mr. Tushar J Bhatt
30
Arithmetic and Geometric Progression
Ex-4: The 4th and 7th terms of a GP are 72 and 576.
Find the sum of first ‘n’ terms
3 6
4 7
6
7
3
4
H e re S u p p o s e th a t th e firs t te rm o f G P i s a n d
c o m m o n ra tio is . H e re g iv e n th a t
7 2 ......(1) a n d 5 7 6 ......( 2 )
T o fin d : a n d
....( 2 )
C o n s id e r w e g e t,
....(1)
5 7 6
8
7 2
a
r
T a r T a r
a r
e q
e q
T a r
T a r
r
   
   

3
8
2 P u t in e q .......(1 ) w e g e t,r

 
Solution 4 :

31. Semester :I
Mr. Tushar J Bhatt
31
Arithmetic and Geometric Progression
3
4
3
(1) 7 2
( 2 ) 7 2
(8 ) 7 2
7 2
8
9
T a r
a
a
a
a
  
 
 
 
 
Solution 4 :
 
 
 
T o fin d :
H e re 2 1 th e n w e a p p ly th e fo rm u la
1
w h e re 9 a n d 2
1
2 1
9
2 1
9 2 1
n
n
n
n
n
n
n
S
r
r
S a a r
r
S
S
 

   


  

   Answer

32. Semester :I
Mr. Tushar J Bhatt
32
Arithmetic and Geometric Progression
Harmonic Progression
Definition : Harmonic Progression(HP)
1 1 1 1
(1) A seq u en ce , , , , .... is in H arm o n ic P ro g ressio n b ecau se
2 4 6 8
2 , 4 , 6 , 8, .... is an arith m etic p ro g ressio n .
For example:
1 2
1 2
A seq u en ce , , ..., is said to b e in H arm o n ic P ro g ressio n
1 1 1
w h en th eir recip ro cal , , ..., are in arith m etic p ro g ressio n .
n
n
x x x
x x x
1 1 1 1
(2 ) A seq u en ce , , , , .... is in H arm o n ic P ro g ressio n b ecau se
5 8 1 1 1 4
5, 8,1 1,1 4, .... is an arith m etic p ro g ress io n .

33. Semester :I
Mr. Tushar J Bhatt
33
Arithmetic and Geometric Progression
1 1 1 1
E X -1 : F in d 2 9 te rm o f th e s e q u e n c e , , , , ....
4 7 1 0 1 3
th
2 9 2 9
1 1 1 1
H ere th e seq u en ce , , , , .... is in H arm o n ic
4 7 1 0 1 3
P ro g ressio n b ecau se a recip ro cal o f th e g iven seq u en ce
4 , 7 , 1 0 , 1 3 ,... is an A P .
T o fin d : o f an A P an d tak e th eir recip ro c al is th e o fT T H P
Solution 1 :
N o w th e g iv e n A P is 4 , 7 , 1 0 , 1 3 ,...
in w h ic h 4 , 3, 2 9
W e k n o w th a t T ( 1)n
a d n
a n d
  
  

34. Semester :I
Mr. Tushar J Bhatt
34
Arithmetic and Geometric Progression
Solution 1 :
2 9
2 9
2 9
2 9
2 9
4 ( 2 9 1) 3
4 ( 2 8 ) 3
4 8 4
8 8 is th e 2 9 te rm o f a n A P
1
T h a t re c ip ro c a l is th e 2 9 te rm o f H P
8 8
1
i.e . 2 9 te rm o f g iv e n H P is .
8 8
th
th
th
T
T
T
T
T
    
   
  
 

 Answer

35. Semester :I
Mr. Tushar J Bhatt
35
Arithmetic and Geometric Progression
Arithmetic Mean
If th e 3 n u m b ers are in A P , th en th e m id d le n u m b er
is said to b e th e arith m etic m ean (A .M ) o f th e first an d th e
th ird n u m b ers.
L et , , are in A P th en th e m id d le n u m b er is said to b e ana A b A A M .
L et ' ' an d ' ' are an y tw o n u m b ers th en arith m etic m ean (A M ) o f
tw o n u m b ers ' ' an d ' 'is o b tain ed b y d iv id in g th e su m o f tw o
n u m b ers b y 2 .
A n arith m etic m ean w e sim p ly d en o te as A .
i.e. A =
2
a b
a b
a b
.

36. Semester :I
Mr. Tushar J Bhatt
36
Arithmetic and Geometric Progression
Geometric Mean
If th e 3 n u m b ers are in G P , th en th e m i d d le n u m b er
is said to b e th e g eo m etric m ean (G .M ) o f th e first an d th e
th ird n u m b ers.
L et , , are in G P th en th e m id d le n u m b er G is said to b e an Ga G b M .
L et ' ' an d ' ' are an y tw o n u m b ers th en g eo m etric m ean (G M ) o f
tw o n u m b ers ' ' an d ' 'is o b tain ed b y tak in g sq u are ro o t o f th e
p ro d u ct o f th e tw o n u m b ers.
T h e g eo m etric m ean w e sim p ly d en o te as G .
a b
a b
i.e. G = .a b

37. Semester :I
Mr. Tushar J Bhatt
37
Arithmetic and Geometric Progression
Example of Arithmetic Mean
F o r e x a m p le le t 3 , 5 , 7 a re in A P
T h e re fo re m id d le n u m b e r(te rm ) 5 b e c o m e a n A M o f
3 a n d 7 . It is a lso o b ta in e d b y u sin g fo rm u la
3 7 1 0
= 5 .
2 2 2
a b
A A
 
   
Example of Geometric Mean
F o r ex am p le let 5 , 1 0 , 2 0 are in G P
T h erefo re m id d le n u m b er(term ) 1 0 b eco m e G M o f
5 an d 2 0 . It is also o b tain ed b y u sin g fo rm u la
5 2 0 1 0 0 1 0 .
( W e co n sid er o n ly p o sitiv e v alu e)
G a b G        

38. Semester :I
Mr. Tushar J Bhatt
38
Arithmetic and Geometric Progression
E X -1 : F in d th e A M a n d G M o f th e fo llo w in g n u m b e rs :
1
(i) 8 a n d 3 2 (ii) 2 a n d 1 8 (iii) a n d 8
3 2
Solution 1 :
( ) H e re 8 a n d 3 2
T o F in d : A M
A n A rith m e tic M e a n
2
8 3 2 4 0
2 0
2 2
T o F in d : G M
T h e G e o m e tric M e a n G
G 8 3 2 2 5 6 1 6
i a b
a b
A
A
a b
 



   
 
      

39. Semester :I
Mr. Tushar J Bhatt
39
Arithmetic and Geometric Progression
Solution 1 :
( ) H e re 2 a n d 1 8
T o F in d : A M
A n A rith m e tic M e a n
2
2 1 8 2 0
1 0
2 2
T o F in d : G M
T h e G e o m e tric M e a n G
G 2 1 8 3 6 6
ii a b
a b
A
A
a b
 



   
 
      

40. Semester :I
Mr. Tushar J Bhatt
40
Arithmetic and Geometric Progression
Solution 1 :
1
( ) H e re a n d 8
3 2
T o F in d : A M
A n A rith m e tic M e a n
2
1 1 3 2 8
8
1 2 5 6 2 5 73 2 3 2
2 2 2 3 2 6 4
T o F in d : G M
T h e G e o m e tric M e a n G
1 1 1
G 8
3 2 4 2
iii a b
a b
A
A
a b
 


 


    

 
      

41. Semester :I
Mr. Tushar J Bhatt
41
Arithmetic and Geometric Progression
E X -2 : T h e A M a n d G M o f th e tw o n u m b e rs a re 2 5 .5 a n d 1 2
re s p e c tiv e ly , fin d th e n u m b e rs .
Solution 2 :
2
S u p p o s e th a t tw o n u m b e r a re ' ' a n d ' '.
th e n 2 5 .5 5 1 ....(1)
2 2
1 4 4
A n d G 1 2 1 4 4 ....( 2 )
1 4 4
P u t in 5 1 w e g e t,
1 4 4
5 1
1 4 4 5 1
a b
a b a b
A a b
a b a b a b b
a
b a b
a
a
a
a a
 
     
        
  
 
  

42. Semester :I
Mr. Tushar J Bhatt
42
Arithmetic and Geometric Progression
Solution 2 :
2
5 1 1 4 4 0a a   
( 4 8)( 3) 0a a   
E ither 48 0 or 3 0a a    
E ith er 4 8 o r 3a a  
1 4 4
N o w w e h a v e b
a

1 4 4
If 4 8 th en 3 . ( , ) (4 8, 3)
4 8
a b i e a b    
1 4 4
If 3 th en 4 8 . ( , ) (3, 4 8).
3
a b i e a b    

43. Semester :I
Mr. Tushar J Bhatt
43
Arithmetic and Geometric Progression
E X -3 : If th e th ree n u m b ers 3, 3 an d 4 are in G P .
F in d th e v alu e o f .
k k
k

Solution 3 :
2
2
2
2
2
2
H e re th re e n u m b e rs 3 , 3, 4 a re in G P .
T h e re fo re th e m id d le te rm 3 re p re s e n t th e G M o f 3 a n d 4 .
. : 3, 3 a n d 4 .
w e k n o w th a t
( 3 ) 3 4
6 9 1 2
6 9 1 2 0
6 9 0
( 3 ) 0
k k
k k
i e G k a b k
G a b
G a b
k k
k k k
k k k
k k
k


   
 
 
   
   
    
   
  
3 0
3 .
k
k
  
  Answer

44. Semester :I
Mr. Tushar J Bhatt
44
Arithmetic and Geometric Progression
Harmonic Mean
If th e 3 n u m b ers are in H P , th en th e m i d d le n u m b er
is said to b e th e h arm o n ic m ean (H .M ) o f th e first an d th e
th ird n u m b ers.
L et , , are in H P th en th e m id d le n u m b er H is said to b e H M .a H b
T h e h arm o n ic m ean w e sim p ly d en o te as H .
L et ' ' an d ' ' are an y tw o n u m b ers th en h arm o n ic m ean (H M ) o f
tw o n u m b ers ' ' an d ' 'is o b tain ed b y u si n g th e fo rm u la
2
H = .
a b
a b
a b
a b

45. Semester :I
Mr. Tushar J Bhatt
45
Arithmetic and Geometric Progression
Relation Between AM, GM and HM
Q u e stio n -1 : P ro v e th a t fo r a n y tw o re a l n u m b e rs th e n
A M G M H M . 
OR
2
P ro o f :
L e t ' ' a n d ' ' a re a n y tw o re a l n u m b e rs t h e n
(1 ) A M = A =
2
( 2 ) G M = G = G =
2
(3 ) H M = H =
a b
a b
a b a b
a b
a b

 


46. Semester :I
Mr. Tushar J Bhatt
46
Arithmetic and Geometric Progression
Relation Between AM, GM and HM
   
    
2
2 2
2 2
P ro o f :
N o w c o n s id e r re s u lt (1 ) re s u lt (3 ) w e g e t,
2
.........( )
2
N o w c o n s id e r re s u lt (1 ) - re s u lt (2 ) w e g e t,
2
22
2 2
,
a b a b
A H a b G i
a b
a b
A G a b
a b a ba b a b
A G
a a b b

   
       
   

  
  
   
 

47. Semester :I
Mr. Tushar J Bhatt
47
Arithmetic and Geometric Progression
Relation Between AM, GM and HM
   
 
 
2 2
2
2 2 2
P ro o f :
2
2
0 ( ) 2
2
0
..............( )
a a b b
A G
a b
A G a b a a b b
A G
A G ii
 
  

       
  
 
2 2
R e su lt ......(ii) B o th sid e s m u ltip lyin g b y H w e g e t,
( .....( ) )
............( )
A H G H
G G H A H G i
G H iii
 
  
 

48. Semester :I
Mr. Tushar J Bhatt
48
Arithmetic and Geometric Progression
Relation Between AM, GM and HM
P ro o f :
F ro m re su lt ..............( ) a n d re s u lt ............( )
W e h a v e
H e n c e p ro v e d th e re su lt.
A G ii G H iii
A G H
 
 

49. Semester :I
Mr. Tushar J Bhatt 49
Arithmetic and Geometric Progression
E X - 4: F ind A M , G M and H M of the num bers 8 and 18.
Solution 4 :
T o Find : A M
W e know that the A M is given by
2
8 18
2
26
2
13
a b
A
A
A
A



 
 
  1st Answer
T o F in d : G M
w e k n o w th at th e G M is g iven b y
8 1 8
1 4 4
1 2 b u t w e co n sid er o n ly
p o sitive valu e th erefo re
1 2
G a b
G
G
G
G
 
   
  
  

H ere ' 8 ' and ' 18 'a b 
2nd Answer
T o F ind : H M
w e know that the H M is given by
2
2 8 18
8 18
288
26
144
13
ab
H
a b
H
H
H


 
 

 
  3rd Answer

50. Semester :I
Mr. Tushar J Bhatt 50
Arithmetic and Geometric Progression
2
E X - 5 : F o r tw o n u m b ers 5 , 4 4 , verify th a t (i) G = A H (ii) A > G > H .
Solution 5 :
T o Find : A M
W e know that the A M is given by
2
5 44
2
49
2
24.5
a b
A
A
A
A



 
 
 
H ere ' 5 ' and ' 44 'a b 
T o Find : H M
w e know that the H M is given by
2
2 5 44
5 44
440
49
8.98
ab
H
a b
H
H
H


 
 

 
  
22
2
T o F in d : G M
w e k n o w th at th e G M is g iven b y
5 4 4
2 2 0
1 4 .8 3 b u t w e co n sid er o n ly
p o sitive valu e th erefo re
1 4 .8 3
1 4 .8 3 2 1 9 .9 2
2 2 0
G a b
G
G
G
G
G
G
 
   
  
  

  
 

51. Semester :I
Mr. Tushar J Bhatt 51
Arithmetic and Geometric Progression
Solution 5 :
2
2
T o V erify : (i) G
R H S =
2 4 .5 8 .9 8
2 2 0 .0 1
2 2 0
A H
A H
G
L H S

 




T o V erify : (ii)
2 4 .5 1 4 .8 3 8 .9 8
A G H
A G H
 
    

52. Semester :I
Mr. Tushar J Bhatt 52
Arithmetic and Geometric Progression
Assignment - 1
Sr.No Instruction Answer
(1) Define arithmetic progression and geometric progression and
give formulae for finding the term and sum of first ‘n’ terms of
these progressions.
(2) Find the required term of the following sequence :
a. 10, 14, 18, 22, …….30th term
b. 16, 26, 36, 46, ……..15th term
c. 59, 56, 53, 50, …….17th term
a. 126
b. 156
c. 011
(3) The 12th term of an arithmetic progression is 20 and its 32th
term is 60, find its 40th term.
76
(4) The 20th term of an arithmetic progression is 30 and its 30th
term is 20, find its 50th term.
0
(5) Find the sum of following AP’s
a. 5, 9, 13, 17, ……( up to 10 terms)
b. 32, 28, 24, 20, …….( up to 13 terms )
c. 1, 3, 5, 7, …..( up to 50 terms )
a. 230
b. 104
c. 2500

53. Semester :I
Mr. Tushar J Bhatt 53
Arithmetic and Geometric Progression
Assignment - 1
Sr.No Instruction Answer
(6) The 3rd term of an AP is 9 and its 9th term is 21, find the sum
of its 40 terms.
1760
(7) The 4th term of an AP is 3 and its 10th term is -9, find the sum
of its 20 terms.
-200
(8) The sum of 10 terms of an AP is 230 and the sum of its 4
terms is 44, find the sum of its 14 terms.
434
(9) The sum of 5 terms of an AP is 25 and the sum of its 9 terms is
81, find the sum of its 10 terms.
100
(10) How many terms of the sequence 2, 5, 8, 11, ……. Will make
the sum 610?
20

54. Semester :I
Mr. Tushar J Bhatt 54
Arithmetic and Geometric Progression
(1) Define arithmetic progression and geometric progression and give
formulae for finding the term and sum of first ‘n’ terms of these progressions.

55. Semester :I
Mr. Tushar J Bhatt 55
Arithmetic and Geometric Progression
(2) Find the required term of the following sequence :
a. 10, 14, 18, 22, …….30th term
b. 16, 26, 36, 46, ……..15th term
c. 59, 56, 53, 50, …….17th term
a. 126
b. 156
c. 011

56. Semester :I
Mr. Tushar J Bhatt 56
Arithmetic and Geometric Progression
(3) The 12th term of an arithmetic progression is 20 and its 32th
term is 60, find its 40th term.
76

57. Semester :I
Mr. Tushar J Bhatt 57
Arithmetic and Geometric Progression
(4) The 20th term of an arithmetic progression is 30 and its 30th
term is 20, find its 50th term.
0

58. Semester :I
Mr. Tushar J Bhatt 58
Arithmetic and Geometric Progression
(5) Find the sum of following AP’s
a. 5, 9, 13, 17, ……( up to 10 terms)
b. 32, 28, 24, 20, …….( up to 13 terms )
c. 1, 3, 5, 7, …..( up to 50 terms )
a. 230
b. 104
c. 2500

59. Semester :I
Mr. Tushar J Bhatt 59
Arithmetic and Geometric Progression
(6) The 3rd term of an AP is 9 and its 9th term is 21, find the sum
of its 40 terms.
1760

60. Semester :I
Mr. Tushar J Bhatt 60
Arithmetic and Geometric Progression
(7) The 4th term of an AP is 3 and its 10th term is -9, find the sum
of its 20 terms.
-200

61. Semester :I
Mr. Tushar J Bhatt 61
Arithmetic and Geometric Progression
(8) The sum of 10 terms of an AP is 230 and the sum of its 4
terms is 44, find the sum of its 14 terms.
434

62. Semester :I
Mr. Tushar J Bhatt 62
Arithmetic and Geometric Progression
(9) The sum of 5 terms of an AP is 25 and the sum of its 9 terms is
81, find the sum of its 10 terms.
100

63. Semester :I
Mr. Tushar J Bhatt 63
Arithmetic and Geometric Progression
(10) How many terms of the sequence 2, 5, 8, 11, ……. Will make
the sum 610?
20

64. Semester :I
Mr. Tushar J Bhatt
64
Arithmetic and Geometric Progression