Arithmetic and Geometric Progressions

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 firs t te rm o f a n y s e q u e c e is d e n o te d b y T ,
T h e s e c o n d te rm is d e n o te d b y T ,
T h e th ird te rm is d e n o te d b y T ,
T h e n te rm is d e n o te d 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 s u m o f firs t 'n ' te rm s o f a s e rie s is d e n o te d b y S ,
T h e re fo 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 te 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)
st
nd
rd
th
n
a
d
ad
ad
ad
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 ted 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 '
st
nd
rd
th
n
a
d
a
ad
ad
n a n d



1 2 3
n ' te rm s o f a n A P is
...
nn
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 s id e r e q ....(1 ) + e q ....(2 ) w e g e t,
2 ( 1) ( 2 ) ... ( 1)
2 2 ( 1) 2 ( 1) ... 2 ( 1)
n
n
n t i m e s
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 by
= 2 ( 1)
2
n
n
S a n d 
Note:
Ex-1: Find the 50
th
term of an AP : 37, 33, 29,
25,… Solution 1 : 50
50
50
50
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
159
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 4
th
term of an AP is 19 and its 12
th
term is 51,
then find its 21
st
term.
Solution 2 : 4
12
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
ad
ad
ad
ad
d
d
a
a
a
  
  



  

  
  
 7

9. Semester :I
Mr. Tushar J Bhatt
9
Arithmetic and Geometric Progression
Solution 2 : 21
21
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 .
st
ad
T a d
T

       

Ex-3 : The 6
th
term of an AP is 47 and its 10
th
term is 75,
then find its 30
th
term.
Solution 3 : 6
10
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
ad
ad
ad
ad
d
  
  



   7 p u t in e q u a tio n ......(1 ) w e g e t,
(1) 5 ( 7 ) 4 7
4 7 3 5
12
d
a
a
a

  
  


10. Semester :I
Mr. Tushar J Bhatt
10
Arithmetic and Geometric Progression
Solution 3: 30
30
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
ad
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) :  
 
 
 

20
20
20
20
20
N o w 2 ( 1)
2
20
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
670
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 , ...
22
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) :  
 

30
30
30
30
30
30
N o w 2 ( 1)
2
3 0 5
2 7 (3 0 1)
22
( 2 9 5 )
1 5 1 4
2
2 8 1 4 5
15
2
173
15
2
2595
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 4
th
term of an AP is 22 and its 10
th
term is 52,
find the sum of its 40 terms.
Solution 5 : 4
10
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
ad
ad
ad
ad
d
d
ad
a
a
a
  
  



  

  
   
  


14. Semester :I
Mr. Tushar J Bhatt
14
Arithmetic and Geometric Progression
Solution 5 :  
 
 
 
 
40
40
40
40
40
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
40
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
4180
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 40
th
term and the sum of first 40 terms of
the sequence 1, 3, 5, 7, …….
Solution 6 : 40
40
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
________________________
ad
ad

        
 
 
40
40
________
T o fin d :
________________________________
w e k n o w th a t 2 ( 1)
2
40
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 :  
 
 

40
40
40
40
40
40
2 1 ( 4 0 1) 2
2
2 0 2 3 9 2
2 0 2 7 8
2 0 8 0
1600
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
21
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
20
th
term. 2
2 3 ,nn  
1
22
1
22
22
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
41
n n n
nn
n
n
n
T S S
S n n S n n
T n n n n
T n n n n
Tn



    
     
     
   20
20
20
20
________________________
T o fin d :
________________________
41
4 ( 2 0 ) 1
8 0 1
81
n
T
Tn
T
T
T
  
  
  

Answer

18. Semester :I
Mr. Tushar J Bhatt
18
Arithmetic and Geometric Progression
Geometric Progression
Definition : Geometric Progression(GP) 23
23
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
ar
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 23
11
1
21
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 =
t h n
n
ar
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 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 , , , , ..., th e n
... ......(1)
E q ....(1 ) m u ltip ly in g b y ' ' b o th s id e s we g e t,
(1)
n
n
n
n
ar
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
234
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
nn
nn
n
nn
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
nn
n
n
n
n
n
n
S r S a a r
S r a r
ar
S if r
r
ar
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,……(9
th
term )
(b) 1024, -512, 256, -128,…… (10
th
term) 1
91
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 ive n b y
2 (3 )
2 (3 )
2 6 5 6 1
13122
th
n
n
ar
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
10
9
10
10
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 ive n b y
1
1024
2
1
1024
2
1024
th
n
n
ar
T a r
T
T
T



       



   



   


   
10
1
512
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) 12
12
12
12
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)
21
4 0 9 6 1
1
4095
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 10
10
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
81
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) : 10
1
81
1024
1
2
S




 10
1023
8
1024
1
2
S




 10
1 0 2 3 2
8
1 0 2 4 1
S

   

 10
1 0 2 3 2
1 2 8 1
S   10
1023
64
S

26. Semester :I
Mr. Tushar J Bhatt
26
Arithmetic and Geometric Progression
Ex-3: The 4
th
term of a GP is 4 and product of 2
nd
and
4
th
terms is 1. Find the 6
th
term and sum of first 6
terms. 3
4 3
2
24
3
24
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 is 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
TT
a r a r
ar
a
r
a
r
   


  



Solution 3 :

27. Semester :I
Mr. Tushar J Bhatt
27
Arithmetic and Geometric Progression 23
32
22
C o m p a re ....(1 ) & ....(2 ) w e h a v e
14
4
4
A n d
1 1 1
( 4 ) 1 6
rr
rr
r
a
r



  
Solution 3 : 6
1
N o w 4 a n d
16
________________________
T o fin d : T
________________________
ra

28. Semester :I
Mr. Tushar J Bhatt
28
Arithmetic and Geometric Progression
Solution 3 : 
5
6
5
6
6
6
T
1
T4
16
1024
T
16
T 6 4
ar

  




1
st
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)
11
nn
nn
S
r
a
S S a
rr


   


29. Semester :I
Mr. Tushar J Bhatt
29
Arithmetic and Geometric Progression
Solution 3 :
2
nd
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
48
4095
48
S
S
S
S

  



  






30. Semester :I
Mr. Tushar J Bhatt
30
Arithmetic and Geometric Progression
Ex-4: The 4
th
and 7
th
terms of a GP are 72 and 576.
Find the sum of first ‘n’ terms 36
47
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 is 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)
576
8
72
a
r
T a r T a r
ar
eq
eq
T ar
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
72
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
21
9
21
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 s e q u e n c e , , , , .... is in H a rm o n ic Pro g re s s io n b e c a u s e
2 4 6 8
2 , 4 , 6 , 8 , .... is a n a rith m e tic p ro g re s s ion.
For example:
12
12
A s e q u e n c e , , ..., is s a id to b e in H a rm on ic P ro g re s s io n
1 1 1
w h e n th e ir re c ip ro c a l , , ..., a re in a rith m e tic p ro g re s s io n .
n
n
x x x
x x x 1 1 1 1
( 2 ) A s e q u e n c e , , , , .... is in H a rm o n ic P ro g re s s io n b e c a u s e
5 8 1 1 1 4
5 , 8 ,1 1,1 4 , .... is a n a rith m e tic p ro g re s sio 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 e re th e s e q u e n c e , , , , .... is in H a rm o nic
4 7 1 0 1 3
P ro g re s s io n b e c a u s e a re c ip ro c a l o f th e g iv e n s e q u e n c e
4 , 7 , 1 0 , 1 3 ,... is a n A P .
T o fin d : o f a n A P a n d ta k e th e ir re c ip ro ca l is th e o f TT HP
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 : 29
29
29
29
29
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
88
1
i.e . 2 9 te rm o f g iv e n H P is .
88
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 e rs a re in A P , th e n th e m id d le n u m b e r
is s a id to b e th e a rith m e tic m e a n (A.M ) o f th e firs t a n d th e
th ird n u m b e rs .
L e t , , a re in A P th e n th e m id d le n u m b e r is s a id to b e a n a A b A A M .
L e t ' ' a n d ' ' a re a n y tw o n u m b e rs th e n a rith m e tic m e a n (A M ) o f
tw o n u m b e rs ' ' a n d ' ' is o b ta in e d b y d ivid in g th e s u m o f tw o
n u m b e rs b y 2 .
A n a rith m e tic m e a n w e s im p ly d e n o te a s A.
i.e . A =
2
ab
ab
ab
.

36. Semester :I
Mr. Tushar J Bhatt
36
Arithmetic and Geometric Progression
Geometric Mean If th e 3 n u m b e rs a re in G P , th e n th e m id d le n u m b e r
is s a id to b e th e g e o m e tric m e a n (G .M ) o f th e firs t a n d th e
th ird n u m b e rs .
L e t , , a re in G P th e n th e m id d le n u m b e r G is s a id to b e a n Ga G b M.
L e t ' ' a n d ' ' a re a n y tw o n u m b e rs th e n g e o m e tric m e a n (G M ) o f
tw o n u m b e rs ' ' a n d ' ' is o b ta in e d b y ta kin g s q u a re ro o t o f th e
p ro d u c t o f th e tw o n u m b e rs .
T h e g e o m e tric m e a n w e s im p ly d e n o te a s G.
ab
ab
i.e . G = .ab

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 an A M o f
3 a n d 7 . It is a ls o o b ta in e d b y u s in g form u la
3 7 1 0
= 5 .
2 2 2
ab
AA

   
Example of Geometric Mean F o r e x a m p le le t 5 , 1 0 , 2 0 a re in G P
T h e re fo re m id d le n u m b e r(te rm ) 1 0 b e c o m e G M o f
5 a n d 2 0 . It is a ls o o b ta in e d b y u s in g fo rm u la
5 2 0 1 0 0 1 0 .
( W e c o n s id e r o n ly p o s itiv e v a lu 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 ing 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
32
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
20
22
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
ab
A
A
ab




   

      

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
10
22
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
ab
A
A
ab




   

      

40. Semester :I
Mr. Tushar J Bhatt
40
Arithmetic and Geometric Progression
Solution 1 : 1
( ) H e re a n d 8
32
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 7
3 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
G8
3 2 4 2
iii a b
ab
A
A
ab






    


      

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 r s ar e 2 5 .5 a n d 1 2
r e s p e c tiv e ly , fin d th e n u m b e r s .
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)
22
144
A n d G 1 2 1 4 4 ....( 2 )
144
P u t in 5 1 w e g e t,
144
51
1 4 4 5 1
ab
a b a b
A a b
a b a b a b b
a
b a b
a
a
a
aa

     
        
  

  

42. Semester :I
Mr. Tushar J Bhatt
42
Arithmetic and Geometric Progression
Solution 2 : 2
5 1 1 4 4 0aa    ( 4 8 )( 3) 0aa    E ith er 4 8 0 o r 3 0aa     E ith er 4 8 o r 3aa   144
N o w w e h a v e b
a
 144
If 4 8 th e n 3 . ( , ) ( 4 8, 3)
48
a b i e a b     144
If 3 th e n 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 re e n u m b e rs 3, 3 a n d 4 a re in G P .
F in d th e v a lu e o f .
kk
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 the 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
kk
kk
i e G k a b k
G a b
G a b
kk
k k k
k k k
kk
k


   


   
   
    
   
  
30
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 e rs a re in H P , th e n th e m id d le n u m b e r
is s a id to b e th e h a rm o n ic m e a n (H .M) o f th e firs t a n d th e
th ird n u m b e rs .
L e t , , a re in H P th e n th e m id d le n u m b e r H is s a id to b e H M .a H b
T h e h a rm o n ic m e a n w e s im p ly d e n o te a s H .
L e t ' ' a n d ' ' a re a n y tw o n u m b e rs th e n h a rm o n ic m e a n (H M ) o f
tw o n u m b e rs ' ' a n d ' ' is o b ta in e d b y u s in g th e fo rm u la
2
H = .
ab
ab
ab
ab

45. Semester :I
Mr. Tushar J Bhatt
45
Arithmetic and Geometric Progression
Relation Between AM, GM and HM Q u e s tio 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 th e n
(1 ) A M = A =
2
( 2 ) G M = G = G =
2
(3 ) H M = H =
ab
ab
a b a b
ab
ab




46. Semester :I
Mr. Tushar J Bhatt
46
Arithmetic and Geometric Progression
Relation Between AM, GM and HM 
  
2
22
22
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 et,
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
2
2
22
,
a b a b
A H a b G i
ab
ab
A G a b
a b a b
a b a b
AG
a a b b

   
    
   
   

  


   


47. Semester :I
Mr. Tushar J Bhatt
47
Arithmetic and Geometric Progression
Relation Between AM, GM and HM  
 
 
22
2
2 2 2
P ro o f :
2
2
0 ( ) 2
2
0
..............( )
a a b b
AG
ab
A G a b a a b b
AG
A G ii

  

       
  
 22
R e s u lt ......(ii) B o th s id e s m u ltip ly in 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 s u lt ..............( ) a n d re su lt ............( )
W e h a v e
H e n c e p ro v e d th e re s u 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 in d A M , G M an d H M o f th e n u m b ers 8 an d 1 8 .
Solution 4 : T o F in d : A M
W e k n o w th a t th e A M is g iv e n b y
2
8 1 8
2
26
2
13
ab
A
A
A
A






1
st
Answer T o F in d : G M
w e k n o w th a t th e G M is g iv e n b y
8 1 8
144
1 2 b u t w e c o n s id e r o n ly
p o s itiv e v a lu e th e re fo re
12
G a b
G
G
G
G

   
  
  
 H ere ' 8 ' and ' 18 'ab 2
nd
Answer T o F in d : H M
w e k n o w th a t th e H M is g iv e n b y
2
2 8 1 8
8 1 8
288
26
144
13
ab
H
ab
H
H
H






 3
rd
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 e rs 5 , 4 4 , v e rify th at (i) G = A H (ii) A > G > H .
Solution 5 : T o F in d : A M
W e k n o w th a t th e A M is g iv e n b y
2
5 4 4
2
49
2
2 4 .5
ab
A
A
A
A





 H ere ' 5 ' and ' 44 'ab T o F in d : H M
w e k n o w th a t th e H M is g iv e n b y
2
2 5 4 4
5 4 4
440
49
8 .9 8
ab
H
ab
H
H
H






  
2
2
2
T o F in d : G M
w e k n o w th a t th e G M is g iv e n b y
5 4 4
220
1 4 .8 3 b u t w e c o n s id e r o n ly
p o s itiv e v a lu e th e re fo re
1 4 .8 3
1 4 .8 3 2 1 9 .9 2
220
G a b
G
G
G
G
G
G

   
  
  

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51. Semester :I
Mr. Tushar J Bhatt
51
Arithmetic and Geometric Progression
Solution 5 : 2
2
T o V e rify : (i) G
R H S =
2 4 .5 8 .9 8
2 2 0 .0 1
220
AH
AH
G
L H S





 T o V e rify : (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, …….30
th
term
b.16, 26, 36, 46, ……..15
th
term
c.59, 56, 53, 50, …….17
th
term
a.126
b.156
c.011
(3) The 12
th
term of an arithmetic progression is 20 and its 32
th
term is 60, find its 40
th
term.
76
(4) The 20
th
term of an arithmetic progression is 30 and its 30
th
term is 20, find its 50
th
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 3
rd
term of an AP is 9 and its 9
th
term is 21, find the sum
of its 40 terms.
1760
(7) The 4
th
term of an AP is 3 and its 10
th
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, …….30
th
term
b.16, 26, 36, 46, ……..15
th
term
c.59, 56, 53, 50, …….17
th
term
a.126
b.156
c.011

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

57. Semester :I
Mr. Tushar J Bhatt
57
Arithmetic and Geometric Progression
(4) The 20
th
term of an arithmetic progression is 30 and its 30
th
term is 20, find its 50
th
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 3
rd
term of an AP is 9 and its 9
th
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 4
th
term of an AP is 3 and its 10
th
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