Detailed Notes: Sequences and Series | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


1. SEQUENCE
1.1 Introduction
A sequence can be defined as an ordered collection of things (usually numbers) or a set of numbers arranged one 
after another. Sometimes, sequence is also referred as progression. The numbers a
1
, 
23
a ,a .....a
n
 are known as terms 
or elements of the sequence. The subscript is the set of positive integers 1, 2, 3..... that indicates the position of the 
term in the sequence. T
n
 is used to denote the n
th
 term.
Some examples of a sequence are as follows:
 0, 7, 26......................., 1, 4, 7, 10......................., 2, 4, 6, 8…………………..
Note: The minimum number terms in a sequence should be 3.
Sequence
3, 5, 7, 9,  .....
1 st term
2nd term
3rd term
4rd term
three dots means
goes on forever (infinite)
(”term’’’’, element’’ or ‘’ member’’’’ mean the same thing)
Figure 3.1
1.2 Finite and Infinite Sequences
A sequence containing a finite number of terms is called a finite sequence. If the sequence contains a infinite 
number of terms, it is known as an infinite sequence. It is infinite in the sense that it never ends. Examples of infinite 
and finite sequences are as follows:
 {1, 2, 3, 4......} is an infinite sequence
 {20, 25, 30, 35....} is an infinite sequence
 {1, 3, 5, 7} is the sequence of the first 4 odd numbers, which is a finite sequence
Page 2


1. SEQUENCE
1.1 Introduction
A sequence can be defined as an ordered collection of things (usually numbers) or a set of numbers arranged one 
after another. Sometimes, sequence is also referred as progression. The numbers a
1
, 
23
a ,a .....a
n
 are known as terms 
or elements of the sequence. The subscript is the set of positive integers 1, 2, 3..... that indicates the position of the 
term in the sequence. T
n
 is used to denote the n
th
 term.
Some examples of a sequence are as follows:
 0, 7, 26......................., 1, 4, 7, 10......................., 2, 4, 6, 8…………………..
Note: The minimum number terms in a sequence should be 3.
Sequence
3, 5, 7, 9,  .....
1 st term
2nd term
3rd term
4rd term
three dots means
goes on forever (infinite)
(”term’’’’, element’’ or ‘’ member’’’’ mean the same thing)
Figure 3.1
1.2 Finite and Infinite Sequences
A sequence containing a finite number of terms is called a finite sequence. If the sequence contains a infinite 
number of terms, it is known as an infinite sequence. It is infinite in the sense that it never ends. Examples of infinite 
and finite sequences are as follows:
 {1, 2, 3, 4......} is an infinite sequence
 {20, 25, 30, 35....} is an infinite sequence
 {1, 3, 5, 7} is the sequence of the first 4 odd numbers, which is a finite sequence
1.3 Rule
A sequence usually has a rule, on the basis of which the terms in the sequence are built up. With the help of this 
rule, we can find any term involved in the sequence. For example, the sequence {3, 5, 7, 9} starts at the number 3 
and jumps 2 every time.
+2 +2
+2
+2
01 23 45 67 89 9
Figure 3.2
As a Formula:
Saying ‘start at the number 3 and jump 2 every time’ is fine, but it does not help to calculate the 10
th
 term or 100
th
 
term or n
th
 term. Hence, we want a formula for the sequence with “n” in it (where n is any term number). What 
would the rule for {3, 5, 7, 9.......} be? First, we can see the sequence goes up 2 every time; hence, we can guess that 
the rule will be something like ‘2 times n’ (where ‘n’ is the term number). Let us test it out.
n Test Rule Term
1 2n = 2 × 1 = 2 3
2 2n = 2 × 2 = 4 5
3 2n = 2 × 3 = 6 7
That nearly worked! But it is less by 1 every time. Let us try changing it to 2n+1.
n Test Rule Term
1 2n + 1 = 2 × 1 + 1 = 3 3
2 2n + 1 = 2 × 2 + 1 = 5 5
3 2n + 1 = 2 × 3 + 1 = 7 7
That Works: Therefore, instead of saying ‘starts at the number 3 and jumps 2 every time,’ we write the expression 
2n + 1. We can now calculate, e.g. the 100
th
 term as 2 × 100 + 1 = 201.
1.4 Notation
The notation T
n
 is used to represent the general term of the sequence. Here, the position of the term in the 
sequence is represented by n. To mention for the ‘5’
th
 term, just write T
5
.
Thus, the rule for {3, 5, 7, 9…} can be written as the following equation: T
n 
= 2n + 1. 
To calculate the 10
th
 term, we can write T
10 
= 2n + 1 = 2 × 10 + 1 = 21
Illustration 1: Find out the first 4 terms of the sequence, {T
n
} = {–1/n}
n
.  (JEE MAIN)
Sol: By substituting n = 1, 2, 3 and 4 in{T
n
} = {–1/n}
n
, we will get the first 4 terms of given sequence.
 T
1 
= (–1/1)
1
 = –1 
 T
2 
= (–1/2)
2
 
= 1/4
 T
3 
= (–1/3)
3 
= –1/27
 T
4 
= (–1/4)
4 
= 1/256
? {T
n
} = {–1, ¼,–1/27, 1/256 ...}
Page 3


1. SEQUENCE
1.1 Introduction
A sequence can be defined as an ordered collection of things (usually numbers) or a set of numbers arranged one 
after another. Sometimes, sequence is also referred as progression. The numbers a
1
, 
23
a ,a .....a
n
 are known as terms 
or elements of the sequence. The subscript is the set of positive integers 1, 2, 3..... that indicates the position of the 
term in the sequence. T
n
 is used to denote the n
th
 term.
Some examples of a sequence are as follows:
 0, 7, 26......................., 1, 4, 7, 10......................., 2, 4, 6, 8…………………..
Note: The minimum number terms in a sequence should be 3.
Sequence
3, 5, 7, 9,  .....
1 st term
2nd term
3rd term
4rd term
three dots means
goes on forever (infinite)
(”term’’’’, element’’ or ‘’ member’’’’ mean the same thing)
Figure 3.1
1.2 Finite and Infinite Sequences
A sequence containing a finite number of terms is called a finite sequence. If the sequence contains a infinite 
number of terms, it is known as an infinite sequence. It is infinite in the sense that it never ends. Examples of infinite 
and finite sequences are as follows:
 {1, 2, 3, 4......} is an infinite sequence
 {20, 25, 30, 35....} is an infinite sequence
 {1, 3, 5, 7} is the sequence of the first 4 odd numbers, which is a finite sequence
1.3 Rule
A sequence usually has a rule, on the basis of which the terms in the sequence are built up. With the help of this 
rule, we can find any term involved in the sequence. For example, the sequence {3, 5, 7, 9} starts at the number 3 
and jumps 2 every time.
+2 +2
+2
+2
01 23 45 67 89 9
Figure 3.2
As a Formula:
Saying ‘start at the number 3 and jump 2 every time’ is fine, but it does not help to calculate the 10
th
 term or 100
th
 
term or n
th
 term. Hence, we want a formula for the sequence with “n” in it (where n is any term number). What 
would the rule for {3, 5, 7, 9.......} be? First, we can see the sequence goes up 2 every time; hence, we can guess that 
the rule will be something like ‘2 times n’ (where ‘n’ is the term number). Let us test it out.
n Test Rule Term
1 2n = 2 × 1 = 2 3
2 2n = 2 × 2 = 4 5
3 2n = 2 × 3 = 6 7
That nearly worked! But it is less by 1 every time. Let us try changing it to 2n+1.
n Test Rule Term
1 2n + 1 = 2 × 1 + 1 = 3 3
2 2n + 1 = 2 × 2 + 1 = 5 5
3 2n + 1 = 2 × 3 + 1 = 7 7
That Works: Therefore, instead of saying ‘starts at the number 3 and jumps 2 every time,’ we write the expression 
2n + 1. We can now calculate, e.g. the 100
th
 term as 2 × 100 + 1 = 201.
1.4 Notation
The notation T
n
 is used to represent the general term of the sequence. Here, the position of the term in the 
sequence is represented by n. To mention for the ‘5’
th
 term, just write T
5
.
Thus, the rule for {3, 5, 7, 9…} can be written as the following equation: T
n 
= 2n + 1. 
To calculate the 10
th
 term, we can write T
10 
= 2n + 1 = 2 × 10 + 1 = 21
Illustration 1: Find out the first 4 terms of the sequence, {T
n
} = {–1/n}
n
.  (JEE MAIN)
Sol: By substituting n = 1, 2, 3 and 4 in{T
n
} = {–1/n}
n
, we will get the first 4 terms of given sequence.
 T
1 
= (–1/1)
1
 = –1 
 T
2 
= (–1/2)
2
 
= 1/4
 T
3 
= (–1/3)
3 
= –1/27
 T
4 
= (–1/4)
4 
= 1/256
? {T
n
} = {–1, ¼,–1/27, 1/256 ...}
Illustration 2: Write the sequence whose n
th
 term is (i) 2
n
 and (ii) log(nx). 
Sol: By substituting n = 1, 2, 3…….., we will get the sequence.
(i)  n
th
 term = 2
n  
 (ii) n
th
 term = a
n 
= log(nx)
 a
1 
= 2
1
, a
2 
= 2
2
.......   a
1
=log(x)
      a
2 
= log(2x), a
n 
= log(nx)
 Sequence ? 2
1
, 2
2
,.........2
n  
Sequence ? log(x), log(2x),................log(nx)
2. SERIES
Series is something that we get from a given sequence by adding all the terms. If we have a sequence as  
T
1
, T
2
, …. ,T
n
, then the series that we get from this sequence is T
1
 + T
2
 +….+T
n
. S
n
 is used to represent the sum of n 
terms. Hence, S
n
 = T
1
 + T
2
 +…. +T
n
3. SIGMA AND PI NOTATIONS
3.1 Sigma Notation
The meaning of the symbol S (sigma) is summation. To find the sum of any sequence, the symbol S (sigma) is used 
before its n
th
 term. For example: 
(i) 
9
n=1
n
?
=1+2+3+.........+9  (ii) 
n
a a
r 1
ror n
=
??
=1
a
+2
a
+3
a
+........+ n
a
(iii) 
5
il
i 1 11 21 3 1 4 1 51
2i 4 2 1 42 2 42 3 42 4 42 5 4
=
+ + + + + +
= + + + +
+ ×+ ×+ ×+ ×+ ×+
?
Properties of S (Sigma)
(i) 
k
il
a
=
?
 = a + a + a… (k times) = ka, where a is a constant.  (ii)  
k k
il il
ai a i
= =
=
??
, where a is a constant.  
(iii) 
n n n
r r r r
r l r l r l
(a b )  a b
= = =
±= ±
? ??
        
(iv) 
i j j i
n n nn
i j i j
i i j j j j i i
0 0 00
a a a a
= = = =
=
? ? ??
3.2 Pi Notation
The symbol ? denotes the product of similar terms. For example: 
(i) 
6
n1
n
=
?
 = 1 × 2 × 3 × 4 × 5 × 6   (ii) 
k
m
n1
n
=
?
 
= 1
m 
× 2
m 
× 3
m 
× 4
m 
× ............. × k
m
(iii) 
k
n1
n
=
?
 = 1 × 2 × 3 × ........ × k = k!
4. ARITHMETIC PROGRESSION
The sequence in which the successive terms maintain a constant difference is known as an arithmetic progression 
(AP). Consider the following sequences:
  a, a + d, a + 2d, a + 3d
    T
1
,   T
2
,    T
3
,   T
4
Page 4


1. SEQUENCE
1.1 Introduction
A sequence can be defined as an ordered collection of things (usually numbers) or a set of numbers arranged one 
after another. Sometimes, sequence is also referred as progression. The numbers a
1
, 
23
a ,a .....a
n
 are known as terms 
or elements of the sequence. The subscript is the set of positive integers 1, 2, 3..... that indicates the position of the 
term in the sequence. T
n
 is used to denote the n
th
 term.
Some examples of a sequence are as follows:
 0, 7, 26......................., 1, 4, 7, 10......................., 2, 4, 6, 8…………………..
Note: The minimum number terms in a sequence should be 3.
Sequence
3, 5, 7, 9,  .....
1 st term
2nd term
3rd term
4rd term
three dots means
goes on forever (infinite)
(”term’’’’, element’’ or ‘’ member’’’’ mean the same thing)
Figure 3.1
1.2 Finite and Infinite Sequences
A sequence containing a finite number of terms is called a finite sequence. If the sequence contains a infinite 
number of terms, it is known as an infinite sequence. It is infinite in the sense that it never ends. Examples of infinite 
and finite sequences are as follows:
 {1, 2, 3, 4......} is an infinite sequence
 {20, 25, 30, 35....} is an infinite sequence
 {1, 3, 5, 7} is the sequence of the first 4 odd numbers, which is a finite sequence
1.3 Rule
A sequence usually has a rule, on the basis of which the terms in the sequence are built up. With the help of this 
rule, we can find any term involved in the sequence. For example, the sequence {3, 5, 7, 9} starts at the number 3 
and jumps 2 every time.
+2 +2
+2
+2
01 23 45 67 89 9
Figure 3.2
As a Formula:
Saying ‘start at the number 3 and jump 2 every time’ is fine, but it does not help to calculate the 10
th
 term or 100
th
 
term or n
th
 term. Hence, we want a formula for the sequence with “n” in it (where n is any term number). What 
would the rule for {3, 5, 7, 9.......} be? First, we can see the sequence goes up 2 every time; hence, we can guess that 
the rule will be something like ‘2 times n’ (where ‘n’ is the term number). Let us test it out.
n Test Rule Term
1 2n = 2 × 1 = 2 3
2 2n = 2 × 2 = 4 5
3 2n = 2 × 3 = 6 7
That nearly worked! But it is less by 1 every time. Let us try changing it to 2n+1.
n Test Rule Term
1 2n + 1 = 2 × 1 + 1 = 3 3
2 2n + 1 = 2 × 2 + 1 = 5 5
3 2n + 1 = 2 × 3 + 1 = 7 7
That Works: Therefore, instead of saying ‘starts at the number 3 and jumps 2 every time,’ we write the expression 
2n + 1. We can now calculate, e.g. the 100
th
 term as 2 × 100 + 1 = 201.
1.4 Notation
The notation T
n
 is used to represent the general term of the sequence. Here, the position of the term in the 
sequence is represented by n. To mention for the ‘5’
th
 term, just write T
5
.
Thus, the rule for {3, 5, 7, 9…} can be written as the following equation: T
n 
= 2n + 1. 
To calculate the 10
th
 term, we can write T
10 
= 2n + 1 = 2 × 10 + 1 = 21
Illustration 1: Find out the first 4 terms of the sequence, {T
n
} = {–1/n}
n
.  (JEE MAIN)
Sol: By substituting n = 1, 2, 3 and 4 in{T
n
} = {–1/n}
n
, we will get the first 4 terms of given sequence.
 T
1 
= (–1/1)
1
 = –1 
 T
2 
= (–1/2)
2
 
= 1/4
 T
3 
= (–1/3)
3 
= –1/27
 T
4 
= (–1/4)
4 
= 1/256
? {T
n
} = {–1, ¼,–1/27, 1/256 ...}
Illustration 2: Write the sequence whose n
th
 term is (i) 2
n
 and (ii) log(nx). 
Sol: By substituting n = 1, 2, 3…….., we will get the sequence.
(i)  n
th
 term = 2
n  
 (ii) n
th
 term = a
n 
= log(nx)
 a
1 
= 2
1
, a
2 
= 2
2
.......   a
1
=log(x)
      a
2 
= log(2x), a
n 
= log(nx)
 Sequence ? 2
1
, 2
2
,.........2
n  
Sequence ? log(x), log(2x),................log(nx)
2. SERIES
Series is something that we get from a given sequence by adding all the terms. If we have a sequence as  
T
1
, T
2
, …. ,T
n
, then the series that we get from this sequence is T
1
 + T
2
 +….+T
n
. S
n
 is used to represent the sum of n 
terms. Hence, S
n
 = T
1
 + T
2
 +…. +T
n
3. SIGMA AND PI NOTATIONS
3.1 Sigma Notation
The meaning of the symbol S (sigma) is summation. To find the sum of any sequence, the symbol S (sigma) is used 
before its n
th
 term. For example: 
(i) 
9
n=1
n
?
=1+2+3+.........+9  (ii) 
n
a a
r 1
ror n
=
??
=1
a
+2
a
+3
a
+........+ n
a
(iii) 
5
il
i 1 11 21 3 1 4 1 51
2i 4 2 1 42 2 42 3 42 4 42 5 4
=
+ + + + + +
= + + + +
+ ×+ ×+ ×+ ×+ ×+
?
Properties of S (Sigma)
(i) 
k
il
a
=
?
 = a + a + a… (k times) = ka, where a is a constant.  (ii)  
k k
il il
ai a i
= =
=
??
, where a is a constant.  
(iii) 
n n n
r r r r
r l r l r l
(a b )  a b
= = =
±= ±
? ??
        
(iv) 
i j j i
n n nn
i j i j
i i j j j j i i
0 0 00
a a a a
= = = =
=
? ? ??
3.2 Pi Notation
The symbol ? denotes the product of similar terms. For example: 
(i) 
6
n1
n
=
?
 = 1 × 2 × 3 × 4 × 5 × 6   (ii) 
k
m
n1
n
=
?
 
= 1
m 
× 2
m 
× 3
m 
× 4
m 
× ............. × k
m
(iii) 
k
n1
n
=
?
 = 1 × 2 × 3 × ........ × k = k!
4. ARITHMETIC PROGRESSION
The sequence in which the successive terms maintain a constant difference is known as an arithmetic progression 
(AP). Consider the following sequences:
  a, a + d, a + 2d, a + 3d
    T
1
,   T
2
,    T
3
,   T
4
	? T
2 
– T
1
 = T
3 
– T
2 
= T
4 
– T
3 
= constant (common difference)
The given sequence is an example of AP . The set of natural numbers is also an example of AP .
4.1 General Term
General term (n
th
 term) of an AP is given by T
n 
= a + (n – 1) d, where a is the first term of the sequence and d is the 
common difference of the sequence.
Note:
(i) General term is also denoted by ? (last term).
(ii) n (number of terms) always belongs to the set of natural numbers.
(iii) Common difference can be zero, + ve or – ve.
  If d > 0 ? increasing AP and the sequence tends to +8
  If d < 0 ? decreasing AP and the sequence tends to -8
  If d = 0 ? constant AP (all the terms remain same)
(iv) The n
th
 term from end is (m – n + 1) term from the beginning, where m is the total number of terms and is given 
by the following expression:
T
m-n+1
 =T
m
 - (n–1) d
•   If the m
th
 term is n and the n
th
 term is m, then the (m + n)
th
 term is 0.
•   If m times the m
th
 term is equal to n times the n
th
 term, then the (m + n)
th
 term is 0.
Vaibhav Krishnan (JEE 2009, AIR 54)
Illustration 3: If the 5
th
 term of an AP is 17 and its 7th term is 15, then find the 22
th
 term. (JEE MAIN)
Sol: Using the formula T
n 
= a + (n – 1) d, we can solve above problem.
 Given a + 4d = 17 and a + 6d = 15
? 2d = -2 ? d = -1, a = 21 
? T
22
 = 21 - 21 = 0
Illustration 4: If 11 times the 11
th
 term of an AP is equal to 9 times the 9
th
 term, then find the 20
th
 term. (JEE MAIN)
Sol: By solving 11 (a + 10d) = 9 (a + 8d), we will get the value of a and d.
? 2a = –38 d ? a = –19d
? 20
th
 term = a + 19d = 0
Illustration 5: Check whether the sequences given below are AP or not.  (JEE MAIN)
(i) T
n
 = n
2  
(ii) T
n
 = an + b
Sol: By taking the difference of two consecutive terms, we can check whether the sequences are in AP or not. 
(i) T
n
 = n
2
; T
n–1
 = (n – 1)
2
 
MASTERJEE CONCEPTS
Page 5


1. SEQUENCE
1.1 Introduction
A sequence can be defined as an ordered collection of things (usually numbers) or a set of numbers arranged one 
after another. Sometimes, sequence is also referred as progression. The numbers a
1
, 
23
a ,a .....a
n
 are known as terms 
or elements of the sequence. The subscript is the set of positive integers 1, 2, 3..... that indicates the position of the 
term in the sequence. T
n
 is used to denote the n
th
 term.
Some examples of a sequence are as follows:
 0, 7, 26......................., 1, 4, 7, 10......................., 2, 4, 6, 8…………………..
Note: The minimum number terms in a sequence should be 3.
Sequence
3, 5, 7, 9,  .....
1 st term
2nd term
3rd term
4rd term
three dots means
goes on forever (infinite)
(”term’’’’, element’’ or ‘’ member’’’’ mean the same thing)
Figure 3.1
1.2 Finite and Infinite Sequences
A sequence containing a finite number of terms is called a finite sequence. If the sequence contains a infinite 
number of terms, it is known as an infinite sequence. It is infinite in the sense that it never ends. Examples of infinite 
and finite sequences are as follows:
 {1, 2, 3, 4......} is an infinite sequence
 {20, 25, 30, 35....} is an infinite sequence
 {1, 3, 5, 7} is the sequence of the first 4 odd numbers, which is a finite sequence
1.3 Rule
A sequence usually has a rule, on the basis of which the terms in the sequence are built up. With the help of this 
rule, we can find any term involved in the sequence. For example, the sequence {3, 5, 7, 9} starts at the number 3 
and jumps 2 every time.
+2 +2
+2
+2
01 23 45 67 89 9
Figure 3.2
As a Formula:
Saying ‘start at the number 3 and jump 2 every time’ is fine, but it does not help to calculate the 10
th
 term or 100
th
 
term or n
th
 term. Hence, we want a formula for the sequence with “n” in it (where n is any term number). What 
would the rule for {3, 5, 7, 9.......} be? First, we can see the sequence goes up 2 every time; hence, we can guess that 
the rule will be something like ‘2 times n’ (where ‘n’ is the term number). Let us test it out.
n Test Rule Term
1 2n = 2 × 1 = 2 3
2 2n = 2 × 2 = 4 5
3 2n = 2 × 3 = 6 7
That nearly worked! But it is less by 1 every time. Let us try changing it to 2n+1.
n Test Rule Term
1 2n + 1 = 2 × 1 + 1 = 3 3
2 2n + 1 = 2 × 2 + 1 = 5 5
3 2n + 1 = 2 × 3 + 1 = 7 7
That Works: Therefore, instead of saying ‘starts at the number 3 and jumps 2 every time,’ we write the expression 
2n + 1. We can now calculate, e.g. the 100
th
 term as 2 × 100 + 1 = 201.
1.4 Notation
The notation T
n
 is used to represent the general term of the sequence. Here, the position of the term in the 
sequence is represented by n. To mention for the ‘5’
th
 term, just write T
5
.
Thus, the rule for {3, 5, 7, 9…} can be written as the following equation: T
n 
= 2n + 1. 
To calculate the 10
th
 term, we can write T
10 
= 2n + 1 = 2 × 10 + 1 = 21
Illustration 1: Find out the first 4 terms of the sequence, {T
n
} = {–1/n}
n
.  (JEE MAIN)
Sol: By substituting n = 1, 2, 3 and 4 in{T
n
} = {–1/n}
n
, we will get the first 4 terms of given sequence.
 T
1 
= (–1/1)
1
 = –1 
 T
2 
= (–1/2)
2
 
= 1/4
 T
3 
= (–1/3)
3 
= –1/27
 T
4 
= (–1/4)
4 
= 1/256
? {T
n
} = {–1, ¼,–1/27, 1/256 ...}
Illustration 2: Write the sequence whose n
th
 term is (i) 2
n
 and (ii) log(nx). 
Sol: By substituting n = 1, 2, 3…….., we will get the sequence.
(i)  n
th
 term = 2
n  
 (ii) n
th
 term = a
n 
= log(nx)
 a
1 
= 2
1
, a
2 
= 2
2
.......   a
1
=log(x)
      a
2 
= log(2x), a
n 
= log(nx)
 Sequence ? 2
1
, 2
2
,.........2
n  
Sequence ? log(x), log(2x),................log(nx)
2. SERIES
Series is something that we get from a given sequence by adding all the terms. If we have a sequence as  
T
1
, T
2
, …. ,T
n
, then the series that we get from this sequence is T
1
 + T
2
 +….+T
n
. S
n
 is used to represent the sum of n 
terms. Hence, S
n
 = T
1
 + T
2
 +…. +T
n
3. SIGMA AND PI NOTATIONS
3.1 Sigma Notation
The meaning of the symbol S (sigma) is summation. To find the sum of any sequence, the symbol S (sigma) is used 
before its n
th
 term. For example: 
(i) 
9
n=1
n
?
=1+2+3+.........+9  (ii) 
n
a a
r 1
ror n
=
??
=1
a
+2
a
+3
a
+........+ n
a
(iii) 
5
il
i 1 11 21 3 1 4 1 51
2i 4 2 1 42 2 42 3 42 4 42 5 4
=
+ + + + + +
= + + + +
+ ×+ ×+ ×+ ×+ ×+
?
Properties of S (Sigma)
(i) 
k
il
a
=
?
 = a + a + a… (k times) = ka, where a is a constant.  (ii)  
k k
il il
ai a i
= =
=
??
, where a is a constant.  
(iii) 
n n n
r r r r
r l r l r l
(a b )  a b
= = =
±= ±
? ??
        
(iv) 
i j j i
n n nn
i j i j
i i j j j j i i
0 0 00
a a a a
= = = =
=
? ? ??
3.2 Pi Notation
The symbol ? denotes the product of similar terms. For example: 
(i) 
6
n1
n
=
?
 = 1 × 2 × 3 × 4 × 5 × 6   (ii) 
k
m
n1
n
=
?
 
= 1
m 
× 2
m 
× 3
m 
× 4
m 
× ............. × k
m
(iii) 
k
n1
n
=
?
 = 1 × 2 × 3 × ........ × k = k!
4. ARITHMETIC PROGRESSION
The sequence in which the successive terms maintain a constant difference is known as an arithmetic progression 
(AP). Consider the following sequences:
  a, a + d, a + 2d, a + 3d
    T
1
,   T
2
,    T
3
,   T
4
	? T
2 
– T
1
 = T
3 
– T
2 
= T
4 
– T
3 
= constant (common difference)
The given sequence is an example of AP . The set of natural numbers is also an example of AP .
4.1 General Term
General term (n
th
 term) of an AP is given by T
n 
= a + (n – 1) d, where a is the first term of the sequence and d is the 
common difference of the sequence.
Note:
(i) General term is also denoted by ? (last term).
(ii) n (number of terms) always belongs to the set of natural numbers.
(iii) Common difference can be zero, + ve or – ve.
  If d > 0 ? increasing AP and the sequence tends to +8
  If d < 0 ? decreasing AP and the sequence tends to -8
  If d = 0 ? constant AP (all the terms remain same)
(iv) The n
th
 term from end is (m – n + 1) term from the beginning, where m is the total number of terms and is given 
by the following expression:
T
m-n+1
 =T
m
 - (n–1) d
•   If the m
th
 term is n and the n
th
 term is m, then the (m + n)
th
 term is 0.
•   If m times the m
th
 term is equal to n times the n
th
 term, then the (m + n)
th
 term is 0.
Vaibhav Krishnan (JEE 2009, AIR 54)
Illustration 3: If the 5
th
 term of an AP is 17 and its 7th term is 15, then find the 22
th
 term. (JEE MAIN)
Sol: Using the formula T
n 
= a + (n – 1) d, we can solve above problem.
 Given a + 4d = 17 and a + 6d = 15
? 2d = -2 ? d = -1, a = 21 
? T
22
 = 21 - 21 = 0
Illustration 4: If 11 times the 11
th
 term of an AP is equal to 9 times the 9
th
 term, then find the 20
th
 term. (JEE MAIN)
Sol: By solving 11 (a + 10d) = 9 (a + 8d), we will get the value of a and d.
? 2a = –38 d ? a = –19d
? 20
th
 term = a + 19d = 0
Illustration 5: Check whether the sequences given below are AP or not.  (JEE MAIN)
(i) T
n
 = n
2  
(ii) T
n
 = an + b
Sol: By taking the difference of two consecutive terms, we can check whether the sequences are in AP or not. 
(i) T
n
 = n
2
; T
n–1
 = (n – 1)
2
 
MASTERJEE CONCEPTS
Difference = T
n 
– T
n–1
 = n
2
 – (n – 1)
2
 = n
2
 – (n
2 
– 2n +1) = 2n – 1
This difference varies with respect to the term. Hence, the sequence is not an AP .
(ii) T
n
 = an + b; T
n – 1
 = a (n – 1) + b
Difference = (an + b) – (a (n – 1) + b) = a (constant)
Hence, the sequence is an AP .
Illustration 6: The 2
nd
, 31
st
 and the last term of an AP are given as 
31
7 ,
42
 and 
1
6
2
- , respectively. Find the first term 
and the number of terms. (JEE MAIN)
Sol: Using T
n 
= a + (n – 1) d, we can get the first term and common difference. Suppose a be the first term and d 
be the common difference of the AP .
Given, 
2
3
T7
4
= ? a+ d = 
31
4
   .... (i)
31
1
T
2
= ? a + 30d = 
1
2
  .... (ii)
Subtracting (i) from (ii), we get  29d = 
1 31 29
– –
24 4
= ? 
1
d
4
-
=
Putting the value of d in (i), we get 
1 31
a–
44
= ? 
31 1 32
a 8
44 4
= + = =
Suppose the number of terms be n, so that T
n 
= 
13
–
2
i.e. a + (n – 1) d = 
13
–
2
 ? 8 + (n – 1) 
1 13
– –
4 2
? ?
=
? ?
? ?
 
? 32 – n + 1 = – 26 ? n = 59 
Hence, the first term = 8 and the number of terms = 59.
Illustration 7: Prove that the square roots of three unequal prime numbers cannot be three terms of an AP . 
  (JEE ADVANCED)
Sol: Here by Considering p , q , r to be the l
th
, µ
th
 and v
th
 terms of an AP and solving them using T
n
 = a + (n – 1) 
d, we prove the problem.
If possible let p , q , r be the three terms of an AP . a, a +d, a + 2d........., where p ? q ? r and they are prime numbers.
Let them be the l
th
, µ
th
 and v
th
 terms, respectively.
? p = a + (l – 1) d
q = a + (µ – 1) d
r = a + (? – 1) d
? p – r = (l – µ) d
Also, q – r = (µ – v) d
? 
p – q –
v
q– r
? µ
=
µ-
 or 
( ) ( )
( ) ( )
p – q q r
–
v
q– r q r
+
? µ
=
µ-
+
 
or 
–
pq pr – q – qr (q – r)
– v
? µ
+ =
µ
 or 
–
pq pr – qr q (q r)
v
? µ
+ =+ -
µ-
 = rational number