Courses

# ARITHMETIC PROGRESSIONS(NCERT) Class 10 Notes | EduRev

## Class 10 : ARITHMETIC PROGRESSIONS(NCERT) Class 10 Notes | EduRev

``` Page 1

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 0
? INTRODUCTION
Consider the following arrangement of numbers :
(i) 1, 3, 5, 7, ....... (ii) 3, 6, 12, 24, ....... (iii) 1, 4, 9, 16, .......
In each of the above arrangements, we observe some patterns. In (i) we find that the succeeding terms are obtained
by adding a fixed number [i.e. 2], in (ii) by multiplying with a fixed number [i.e. 2], in (iii) we find that they
are squares of natural numbers.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed
number to the preceding terms. We shall also see how to find their n
th
terms and the sum of n consecutive
terms, and use this knowledge in solving some daily life problems.
? HISTORICAL FACTS
Gauss was a very talented and gifted mathematician of 19th century who developed the formula :
1 + 2 + 3 + 4 +.... + (n – 1) + n =
n(n 1)
2
?
for the sum of first n natural numbers at the age of 10.
He did this in the following way :
S = 1 + 2 + 3 + ... + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 2) +.....+ 3 + 2 + 1
? 2S = (n + 1) + (n + 1) + (n + 1) +....+ (n + 1) + (n + 1) + (n + 1)
= (n + 1) (1 + 1 + 1 +... upto n times)
2S = (n + 1) n  ? S =
n(n 1)
2
?
Even when he was a little child of three he could read and make mathematical calculation himself. Gauss proved
the fundamental theorem of Algebra when he was 20 years old. His contribution to mathematics has been immense
because his formulae were used in applied field of Astronomy, Differential Geometry and Electricity widely all
over the world by scientists.
? SEQUENCE
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of
students in a row as per their roll numbers, arrangement of books in the library, etc.
An arrangement of numbers depends on the given rule :
Given Rule Arrangement of numbers
Write 3 and then add 4 successively 3,7,11,15 ,19,...
Write 3 and then multiply 4 successively 3, 12, 48, 192,...
Write 4 and then subtract 3 successively 4,1, –2, –5,...
Write alternately 5 and – 5 5, – 5, 5, –5,...
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a Sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth and fifth,.... terms
of the sequence.
The terms of a sequence in successive order is denoted by 'T'
n
or 'a'
n
. The nth term 'T'
n
is called the general
term of the sequence.
ARITHMETIC PROGRESSIONS
ARITHMETIC PROGRESSIONS
Page 2

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 0
? INTRODUCTION
Consider the following arrangement of numbers :
(i) 1, 3, 5, 7, ....... (ii) 3, 6, 12, 24, ....... (iii) 1, 4, 9, 16, .......
In each of the above arrangements, we observe some patterns. In (i) we find that the succeeding terms are obtained
by adding a fixed number [i.e. 2], in (ii) by multiplying with a fixed number [i.e. 2], in (iii) we find that they
are squares of natural numbers.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed
number to the preceding terms. We shall also see how to find their n
th
terms and the sum of n consecutive
terms, and use this knowledge in solving some daily life problems.
? HISTORICAL FACTS
Gauss was a very talented and gifted mathematician of 19th century who developed the formula :
1 + 2 + 3 + 4 +.... + (n – 1) + n =
n(n 1)
2
?
for the sum of first n natural numbers at the age of 10.
He did this in the following way :
S = 1 + 2 + 3 + ... + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 2) +.....+ 3 + 2 + 1
? 2S = (n + 1) + (n + 1) + (n + 1) +....+ (n + 1) + (n + 1) + (n + 1)
= (n + 1) (1 + 1 + 1 +... upto n times)
2S = (n + 1) n  ? S =
n(n 1)
2
?
Even when he was a little child of three he could read and make mathematical calculation himself. Gauss proved
the fundamental theorem of Algebra when he was 20 years old. His contribution to mathematics has been immense
because his formulae were used in applied field of Astronomy, Differential Geometry and Electricity widely all
over the world by scientists.
? SEQUENCE
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of
students in a row as per their roll numbers, arrangement of books in the library, etc.
An arrangement of numbers depends on the given rule :
Given Rule Arrangement of numbers
Write 3 and then add 4 successively 3,7,11,15 ,19,...
Write 3 and then multiply 4 successively 3, 12, 48, 192,...
Write 4 and then subtract 3 successively 4,1, –2, –5,...
Write alternately 5 and – 5 5, – 5, 5, –5,...
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a Sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth and fifth,.... terms
of the sequence.
The terms of a sequence in successive order is denoted by 'T'
n
or 'a'
n
. The nth term 'T'
n
is called the general
term of the sequence.
ARITHMETIC PROGRESSIONS
ARITHMETIC PROGRESSIONS
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
2
? SERIES
The sum of terms of a sequence is called the series of the corresponding sequence. T
1
+ T
2
+ T
3
+.... is an
infinite series, where as T
1
+ T
2
+ T
3
+ ... + T
n–1
+ T
n
is a finite series of n terms.
Usually the series of finite number of n terms is denoted by S
n
.
S
n
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
+ T
n
S
n–1
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
S
n
– S
n–1
= T
n
OR
n n n –1
T S – S ?
Ex.1 Write the first five terms of the sequence, whose nth term is a
n
= {1 + (–1)
n
}n.
Sol. a
n
= {1 + (–1)
n
}n
Substituting n = 1, 2, 3, 4 and 5, we get
a
1
= {1 + (–1)
1
} 1 = 0 ; a
2
= {1 + (–1)
2
} 2 = 4;
a
3
= {1 + (–1)
3
} 3 = 0 ; a
4
= {1 + (–1)
4
} 4 = 8;
a
5
= {1 + (–1)
5
} 5 = 0
Thus, the required terms are : 0, 4, 0, 8 and 0.
Ex.2 Find the 20th term of the sequence whose nth term is, a
n
=
n(n – 2)
n 3 ?
Sol. a
n
=
n(n – 2)
n 3 ?
. Putting n = 20, we obtain a
20
=
20(20 – 2)
20 3 ?
Thus, a
20
=
360
23
Ex.3 The fibonacci sequence is defined by a
1
= 1 = a
2
; a
n
= a
n–1
+ a
n–2
for n > 2. Find
n 1
n
a
a
?
, for n = 1, 2,
3, 4, 5.
Sol. We have a
1
= a
2
= 1 and a
n
= a
n–1
+ a
n–2
Substituting n = 3, 4, 5 and 6, we get.
a
3
= a
2
+ a
1
= 1 + 1 = 2
a
4
= a
3
+ a
2
= 2 + 1 = 3
a
5
= a
4
+ a
3
= 3 + 2 = 5
and a
6
= a
5
+ a
4
= 5 + 3 = 8
Now, we have to find
n 1
n
a
a
?
for n = 1, 2, 3, 4 and 5
For, n = 1,
2
1
a 1
a 1
?
= 1
n = 2,
3
2
a 2
a 1
?
= 2
n = 3,
4
3
a 3
a 2
?
n = 4,
5
4
a 5
a 3
?
n = 5,
6
5
a 8
a 5
?
Hence, the required values are 1,
3 5 8
2, , and
2 3 5
Page 3

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 0
? INTRODUCTION
Consider the following arrangement of numbers :
(i) 1, 3, 5, 7, ....... (ii) 3, 6, 12, 24, ....... (iii) 1, 4, 9, 16, .......
In each of the above arrangements, we observe some patterns. In (i) we find that the succeeding terms are obtained
by adding a fixed number [i.e. 2], in (ii) by multiplying with a fixed number [i.e. 2], in (iii) we find that they
are squares of natural numbers.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed
number to the preceding terms. We shall also see how to find their n
th
terms and the sum of n consecutive
terms, and use this knowledge in solving some daily life problems.
? HISTORICAL FACTS
Gauss was a very talented and gifted mathematician of 19th century who developed the formula :
1 + 2 + 3 + 4 +.... + (n – 1) + n =
n(n 1)
2
?
for the sum of first n natural numbers at the age of 10.
He did this in the following way :
S = 1 + 2 + 3 + ... + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 2) +.....+ 3 + 2 + 1
? 2S = (n + 1) + (n + 1) + (n + 1) +....+ (n + 1) + (n + 1) + (n + 1)
= (n + 1) (1 + 1 + 1 +... upto n times)
2S = (n + 1) n  ? S =
n(n 1)
2
?
Even when he was a little child of three he could read and make mathematical calculation himself. Gauss proved
the fundamental theorem of Algebra when he was 20 years old. His contribution to mathematics has been immense
because his formulae were used in applied field of Astronomy, Differential Geometry and Electricity widely all
over the world by scientists.
? SEQUENCE
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of
students in a row as per their roll numbers, arrangement of books in the library, etc.
An arrangement of numbers depends on the given rule :
Given Rule Arrangement of numbers
Write 3 and then add 4 successively 3,7,11,15 ,19,...
Write 3 and then multiply 4 successively 3, 12, 48, 192,...
Write 4 and then subtract 3 successively 4,1, –2, –5,...
Write alternately 5 and – 5 5, – 5, 5, –5,...
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a Sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth and fifth,.... terms
of the sequence.
The terms of a sequence in successive order is denoted by 'T'
n
or 'a'
n
. The nth term 'T'
n
is called the general
term of the sequence.
ARITHMETIC PROGRESSIONS
ARITHMETIC PROGRESSIONS
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
2
? SERIES
The sum of terms of a sequence is called the series of the corresponding sequence. T
1
+ T
2
+ T
3
+.... is an
infinite series, where as T
1
+ T
2
+ T
3
+ ... + T
n–1
+ T
n
is a finite series of n terms.
Usually the series of finite number of n terms is denoted by S
n
.
S
n
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
+ T
n
S
n–1
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
S
n
– S
n–1
= T
n
OR
n n n –1
T S – S ?
Ex.1 Write the first five terms of the sequence, whose nth term is a
n
= {1 + (–1)
n
}n.
Sol. a
n
= {1 + (–1)
n
}n
Substituting n = 1, 2, 3, 4 and 5, we get
a
1
= {1 + (–1)
1
} 1 = 0 ; a
2
= {1 + (–1)
2
} 2 = 4;
a
3
= {1 + (–1)
3
} 3 = 0 ; a
4
= {1 + (–1)
4
} 4 = 8;
a
5
= {1 + (–1)
5
} 5 = 0
Thus, the required terms are : 0, 4, 0, 8 and 0.
Ex.2 Find the 20th term of the sequence whose nth term is, a
n
=
n(n – 2)
n 3 ?
Sol. a
n
=
n(n – 2)
n 3 ?
. Putting n = 20, we obtain a
20
=
20(20 – 2)
20 3 ?
Thus, a
20
=
360
23
Ex.3 The fibonacci sequence is defined by a
1
= 1 = a
2
; a
n
= a
n–1
+ a
n–2
for n > 2. Find
n 1
n
a
a
?
, for n = 1, 2,
3, 4, 5.
Sol. We have a
1
= a
2
= 1 and a
n
= a
n–1
+ a
n–2
Substituting n = 3, 4, 5 and 6, we get.
a
3
= a
2
+ a
1
= 1 + 1 = 2
a
4
= a
3
+ a
2
= 2 + 1 = 3
a
5
= a
4
+ a
3
= 3 + 2 = 5
and a
6
= a
5
+ a
4
= 5 + 3 = 8
Now, we have to find
n 1
n
a
a
?
for n = 1, 2, 3, 4 and 5
For, n = 1,
2
1
a 1
a 1
?
= 1
n = 2,
3
2
a 2
a 1
?
= 2
n = 3,
4
3
a 3
a 2
?
n = 4,
5
4
a 5
a 3
?
n = 5,
6
5
a 8
a 5
?
Hence, the required values are 1,
3 5 8
2, , and
2 3 5
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 2
COM PE T I T ION W I NDOW
SERIES OF NATURAL NUMBERS
1. The sum of first n natural numbers i.e. 1 + 2 + 3 +....+ n is usually written as ?n.
n(n 1)
n
2
?
?
?
2. The sum of squares of first n natural numbers  i.e. 1
2
+ 2
2
+ 3
2
+ ... + n
2
is usually written as ?n
2
.
2
n(n 1)(2n 1)
n
6
? ?
?
?
3. The sum of cubes of first n natural numbers i.e. 1
3
+ 2
3
+ 3
3
+ ... + n
3
is usually written as ?n
3
.
2
3 2
n(n 1)
n ( n)
2
? ? ?
? ? ?
? ?
? ?
?
? PROGRESSION
It is not always possible to write each and every sequence of some rule.
For example : Sequence of prime numbers 2, 3, 5, 7, 11,... cannot be expressed explicitly by stating a rule
and we do not have any expression for writting the general term of this sequence.
The sequence that follows a certain pattern is called a progression. Thus, the sequence 2, 3, 5, 7, 11,.... is
not a progression. In a progression, we can always write the nth term.
Consider the following collection of numbers : (i) 1, 3, 5, 7...     (ii)
1 1 1 1
, , , ,...
2 3 4 5
From the above collection of numbers, we observe that
( i ) Each term is greater than the previous by 2.
( i i ) In each term the numerator is 1 and the denominator is obtained by adding 1 to the preceding denominator.
Thus, we observe that the collection of numbers given in (i) and (ii) follow a certain pattern and as such are
all progressions.
? ARITHMETIC PROGRESSIONS
An arithmetic progression is that list of numbers in which the first term is given and each term, other
than the first term is obtained by adding a fixed number 'd' to the preceding term.
The fixed term 'd' is known as the common difference of the arithmetic progression. It's value can be positive,
negative or zero. The first term is denoted by 'a' or 'a
1
' and the last term by ' ?'.
Ex. Consider a sequence 6, 10, 14, 18, 22, .....
Here, a
1
= 6, a
2
= 10, a
3
= 14, a
4
= 18, a
5
= 22
a
2
– a
1
= 10 – 6 = 4
a
3
– a
2
= 14 – 10 = 4
a
4
– a
3
= 18 – 14 = 4
- - - - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - -
Therefore, the sequence is an arithmetic progression in which the first term a = 6 and the common difference d = 4.
Symbolical form : Let us denote the first term of an AP by a
1
, second term by a
2
, .... nth term by a
n
and
the common difference by d. Then the AP becomes a
1
, a
2
, a
3
,....,a
n
.
So, a
2
– a
1
= a
3
– a
2
= ... = a
n
– a
n–1
= d.
Page 4

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 0
? INTRODUCTION
Consider the following arrangement of numbers :
(i) 1, 3, 5, 7, ....... (ii) 3, 6, 12, 24, ....... (iii) 1, 4, 9, 16, .......
In each of the above arrangements, we observe some patterns. In (i) we find that the succeeding terms are obtained
by adding a fixed number [i.e. 2], in (ii) by multiplying with a fixed number [i.e. 2], in (iii) we find that they
are squares of natural numbers.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed
number to the preceding terms. We shall also see how to find their n
th
terms and the sum of n consecutive
terms, and use this knowledge in solving some daily life problems.
? HISTORICAL FACTS
Gauss was a very talented and gifted mathematician of 19th century who developed the formula :
1 + 2 + 3 + 4 +.... + (n – 1) + n =
n(n 1)
2
?
for the sum of first n natural numbers at the age of 10.
He did this in the following way :
S = 1 + 2 + 3 + ... + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 2) +.....+ 3 + 2 + 1
? 2S = (n + 1) + (n + 1) + (n + 1) +....+ (n + 1) + (n + 1) + (n + 1)
= (n + 1) (1 + 1 + 1 +... upto n times)
2S = (n + 1) n  ? S =
n(n 1)
2
?
Even when he was a little child of three he could read and make mathematical calculation himself. Gauss proved
the fundamental theorem of Algebra when he was 20 years old. His contribution to mathematics has been immense
because his formulae were used in applied field of Astronomy, Differential Geometry and Electricity widely all
over the world by scientists.
? SEQUENCE
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of
students in a row as per their roll numbers, arrangement of books in the library, etc.
An arrangement of numbers depends on the given rule :
Given Rule Arrangement of numbers
Write 3 and then add 4 successively 3,7,11,15 ,19,...
Write 3 and then multiply 4 successively 3, 12, 48, 192,...
Write 4 and then subtract 3 successively 4,1, –2, –5,...
Write alternately 5 and – 5 5, – 5, 5, –5,...
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a Sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth and fifth,.... terms
of the sequence.
The terms of a sequence in successive order is denoted by 'T'
n
or 'a'
n
. The nth term 'T'
n
is called the general
term of the sequence.
ARITHMETIC PROGRESSIONS
ARITHMETIC PROGRESSIONS
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
2
? SERIES
The sum of terms of a sequence is called the series of the corresponding sequence. T
1
+ T
2
+ T
3
+.... is an
infinite series, where as T
1
+ T
2
+ T
3
+ ... + T
n–1
+ T
n
is a finite series of n terms.
Usually the series of finite number of n terms is denoted by S
n
.
S
n
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
+ T
n
S
n–1
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
S
n
– S
n–1
= T
n
OR
n n n –1
T S – S ?
Ex.1 Write the first five terms of the sequence, whose nth term is a
n
= {1 + (–1)
n
}n.
Sol. a
n
= {1 + (–1)
n
}n
Substituting n = 1, 2, 3, 4 and 5, we get
a
1
= {1 + (–1)
1
} 1 = 0 ; a
2
= {1 + (–1)
2
} 2 = 4;
a
3
= {1 + (–1)
3
} 3 = 0 ; a
4
= {1 + (–1)
4
} 4 = 8;
a
5
= {1 + (–1)
5
} 5 = 0
Thus, the required terms are : 0, 4, 0, 8 and 0.
Ex.2 Find the 20th term of the sequence whose nth term is, a
n
=
n(n – 2)
n 3 ?
Sol. a
n
=
n(n – 2)
n 3 ?
. Putting n = 20, we obtain a
20
=
20(20 – 2)
20 3 ?
Thus, a
20
=
360
23
Ex.3 The fibonacci sequence is defined by a
1
= 1 = a
2
; a
n
= a
n–1
+ a
n–2
for n > 2. Find
n 1
n
a
a
?
, for n = 1, 2,
3, 4, 5.
Sol. We have a
1
= a
2
= 1 and a
n
= a
n–1
+ a
n–2
Substituting n = 3, 4, 5 and 6, we get.
a
3
= a
2
+ a
1
= 1 + 1 = 2
a
4
= a
3
+ a
2
= 2 + 1 = 3
a
5
= a
4
+ a
3
= 3 + 2 = 5
and a
6
= a
5
+ a
4
= 5 + 3 = 8
Now, we have to find
n 1
n
a
a
?
for n = 1, 2, 3, 4 and 5
For, n = 1,
2
1
a 1
a 1
?
= 1
n = 2,
3
2
a 2
a 1
?
= 2
n = 3,
4
3
a 3
a 2
?
n = 4,
5
4
a 5
a 3
?
n = 5,
6
5
a 8
a 5
?
Hence, the required values are 1,
3 5 8
2, , and
2 3 5
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 2
COM PE T I T ION W I NDOW
SERIES OF NATURAL NUMBERS
1. The sum of first n natural numbers i.e. 1 + 2 + 3 +....+ n is usually written as ?n.
n(n 1)
n
2
?
?
?
2. The sum of squares of first n natural numbers  i.e. 1
2
+ 2
2
+ 3
2
+ ... + n
2
is usually written as ?n
2
.
2
n(n 1)(2n 1)
n
6
? ?
?
?
3. The sum of cubes of first n natural numbers i.e. 1
3
+ 2
3
+ 3
3
+ ... + n
3
is usually written as ?n
3
.
2
3 2
n(n 1)
n ( n)
2
? ? ?
? ? ?
? ?
? ?
?
? PROGRESSION
It is not always possible to write each and every sequence of some rule.
For example : Sequence of prime numbers 2, 3, 5, 7, 11,... cannot be expressed explicitly by stating a rule
and we do not have any expression for writting the general term of this sequence.
The sequence that follows a certain pattern is called a progression. Thus, the sequence 2, 3, 5, 7, 11,.... is
not a progression. In a progression, we can always write the nth term.
Consider the following collection of numbers : (i) 1, 3, 5, 7...     (ii)
1 1 1 1
, , , ,...
2 3 4 5
From the above collection of numbers, we observe that
( i ) Each term is greater than the previous by 2.
( i i ) In each term the numerator is 1 and the denominator is obtained by adding 1 to the preceding denominator.
Thus, we observe that the collection of numbers given in (i) and (ii) follow a certain pattern and as such are
all progressions.
? ARITHMETIC PROGRESSIONS
An arithmetic progression is that list of numbers in which the first term is given and each term, other
than the first term is obtained by adding a fixed number 'd' to the preceding term.
The fixed term 'd' is known as the common difference of the arithmetic progression. It's value can be positive,
negative or zero. The first term is denoted by 'a' or 'a
1
' and the last term by ' ?'.
Ex. Consider a sequence 6, 10, 14, 18, 22, .....
Here, a
1
= 6, a
2
= 10, a
3
= 14, a
4
= 18, a
5
= 22
a
2
– a
1
= 10 – 6 = 4
a
3
– a
2
= 14 – 10 = 4
a
4
– a
3
= 18 – 14 = 4
- - - - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - -
Therefore, the sequence is an arithmetic progression in which the first term a = 6 and the common difference d = 4.
Symbolical form : Let us denote the first term of an AP by a
1
, second term by a
2
, .... nth term by a
n
and
the common difference by d. Then the AP becomes a
1
, a
2
, a
3
,....,a
n
.
So, a
2
– a
1
= a
3
– a
2
= ... = a
n
– a
n–1
= d.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
4
General form : In general form, an arithmetic progression with first term 'a' and common difference 'd' can
be represented as follows :
a, a + d, a + 2d, a + 3d, a + 4d,....
Finite AP : An AP in which there are only a finite number of terms is called a finite AP. It may
be noted that each such AP has a last term.
Ex. ( a ) The heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,
148, 149,..., 157.
( b ) The minimum temperatures (in degree Celsius) recorded for a week in the month of January in a city arranged
in ascending order are –3.1, – 3.0, –2.9, – 2.8, – 2.7, – 2.6, – 2.5
Infinite AP : An AP in which the number of terms is not finite is called infinite AP. It is note worthy
that such APs do not have a last term.
Ex. ( a ) 1, 2, 3, 4,......
( b ) 100, 70, 40, 10,.....
Least Information Required : To know about an AP, the minimum information we need to know is to know
both – the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3, then AP is 6, 9, 12, 15,...
Similarly, when
a = – 7, d = – 2, the AP is –7, –9, –11, –13,...
a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3,...
So if we know what a and d are we can list the AP.
Ex.4 In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
( i ) The taxi fare after each km  when the fare is Rs. 15 for the first km and Rs 8 for each additional km.
( i i ) The amount of air present in a cylinder when a vacuum pump removes
1
4
of the air remaining in the
cylinder at a time.
( i i i ) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and
rises by Rs. 50 for each subsequent metre.
( i v ) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at
8% per annum. [NCERT]
Sol. ( i ) Taxi fare for 1 km = Rs. 15 = a
1
Taxi fare for 2 kms = Rs. 15 + 8 = Rs. 23 = a
2
Taxi fare for 3 kms = Rs. 23 + 8 = Rs. 31 = a
3
Taxi fare for 4 kms = Rs. 31 + 8 = Rs. 39 = a
4
and so on.
a
2
– a
1
= Rs. 23 – 15 = Rs. 8
a
3
– a
2
= Rs. 31 – 23 = Rs. 8
a
4
– a
3
= Rs. 39 – 31 = Rs. 8
i.e., a
k+1
– a
k
is the same everytime.
So, this list of numbers form an arithmetic progression with the first term a = Rs 15 and the common
difference d = Rs. 8.
( i i ) Amount of air present in the cylinder = x units (say) = a
1
Amount of air present in the cylinder after one time removal of air by the vacuum pump = x –
x
4
=
3 x
4
units = a
2
Page 5

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 0
? INTRODUCTION
Consider the following arrangement of numbers :
(i) 1, 3, 5, 7, ....... (ii) 3, 6, 12, 24, ....... (iii) 1, 4, 9, 16, .......
In each of the above arrangements, we observe some patterns. In (i) we find that the succeeding terms are obtained
by adding a fixed number [i.e. 2], in (ii) by multiplying with a fixed number [i.e. 2], in (iii) we find that they
are squares of natural numbers.
In this chapter, we shall discuss one of these patterns in which succeeding terms are obtained by adding a fixed
number to the preceding terms. We shall also see how to find their n
th
terms and the sum of n consecutive
terms, and use this knowledge in solving some daily life problems.
? HISTORICAL FACTS
Gauss was a very talented and gifted mathematician of 19th century who developed the formula :
1 + 2 + 3 + 4 +.... + (n – 1) + n =
n(n 1)
2
?
for the sum of first n natural numbers at the age of 10.
He did this in the following way :
S = 1 + 2 + 3 + ... + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 2) +.....+ 3 + 2 + 1
? 2S = (n + 1) + (n + 1) + (n + 1) +....+ (n + 1) + (n + 1) + (n + 1)
= (n + 1) (1 + 1 + 1 +... upto n times)
2S = (n + 1) n  ? S =
n(n 1)
2
?
Even when he was a little child of three he could read and make mathematical calculation himself. Gauss proved
the fundamental theorem of Algebra when he was 20 years old. His contribution to mathematics has been immense
because his formulae were used in applied field of Astronomy, Differential Geometry and Electricity widely all
over the world by scientists.
? SEQUENCE
In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of
students in a row as per their roll numbers, arrangement of books in the library, etc.
An arrangement of numbers depends on the given rule :
Given Rule Arrangement of numbers
Write 3 and then add 4 successively 3,7,11,15 ,19,...
Write 3 and then multiply 4 successively 3, 12, 48, 192,...
Write 4 and then subtract 3 successively 4,1, –2, –5,...
Write alternately 5 and – 5 5, – 5, 5, –5,...
Thus, a sequence is an ordered arrangement of numbers according to a given rule.
Terms of a Sequence : The individual numbers that form a sequence are the terms of a sequence.
For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth and fifth,.... terms
of the sequence.
The terms of a sequence in successive order is denoted by 'T'
n
or 'a'
n
. The nth term 'T'
n
is called the general
term of the sequence.
ARITHMETIC PROGRESSIONS
ARITHMETIC PROGRESSIONS
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
2
? SERIES
The sum of terms of a sequence is called the series of the corresponding sequence. T
1
+ T
2
+ T
3
+.... is an
infinite series, where as T
1
+ T
2
+ T
3
+ ... + T
n–1
+ T
n
is a finite series of n terms.
Usually the series of finite number of n terms is denoted by S
n
.
S
n
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
+ T
n
S
n–1
= T
1
+ T
2
+ T
3
+ ... + T
n – 2
+ T
n – 1
S
n
– S
n–1
= T
n
OR
n n n –1
T S – S ?
Ex.1 Write the first five terms of the sequence, whose nth term is a
n
= {1 + (–1)
n
}n.
Sol. a
n
= {1 + (–1)
n
}n
Substituting n = 1, 2, 3, 4 and 5, we get
a
1
= {1 + (–1)
1
} 1 = 0 ; a
2
= {1 + (–1)
2
} 2 = 4;
a
3
= {1 + (–1)
3
} 3 = 0 ; a
4
= {1 + (–1)
4
} 4 = 8;
a
5
= {1 + (–1)
5
} 5 = 0
Thus, the required terms are : 0, 4, 0, 8 and 0.
Ex.2 Find the 20th term of the sequence whose nth term is, a
n
=
n(n – 2)
n 3 ?
Sol. a
n
=
n(n – 2)
n 3 ?
. Putting n = 20, we obtain a
20
=
20(20 – 2)
20 3 ?
Thus, a
20
=
360
23
Ex.3 The fibonacci sequence is defined by a
1
= 1 = a
2
; a
n
= a
n–1
+ a
n–2
for n > 2. Find
n 1
n
a
a
?
, for n = 1, 2,
3, 4, 5.
Sol. We have a
1
= a
2
= 1 and a
n
= a
n–1
+ a
n–2
Substituting n = 3, 4, 5 and 6, we get.
a
3
= a
2
+ a
1
= 1 + 1 = 2
a
4
= a
3
+ a
2
= 2 + 1 = 3
a
5
= a
4
+ a
3
= 3 + 2 = 5
and a
6
= a
5
+ a
4
= 5 + 3 = 8
Now, we have to find
n 1
n
a
a
?
for n = 1, 2, 3, 4 and 5
For, n = 1,
2
1
a 1
a 1
?
= 1
n = 2,
3
2
a 2
a 1
?
= 2
n = 3,
4
3
a 3
a 2
?
n = 4,
5
4
a 5
a 3
?
n = 5,
6
5
a 8
a 5
?
Hence, the required values are 1,
3 5 8
2, , and
2 3 5
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 2
COM PE T I T ION W I NDOW
SERIES OF NATURAL NUMBERS
1. The sum of first n natural numbers i.e. 1 + 2 + 3 +....+ n is usually written as ?n.
n(n 1)
n
2
?
?
?
2. The sum of squares of first n natural numbers  i.e. 1
2
+ 2
2
+ 3
2
+ ... + n
2
is usually written as ?n
2
.
2
n(n 1)(2n 1)
n
6
? ?
?
?
3. The sum of cubes of first n natural numbers i.e. 1
3
+ 2
3
+ 3
3
+ ... + n
3
is usually written as ?n
3
.
2
3 2
n(n 1)
n ( n)
2
? ? ?
? ? ?
? ?
? ?
?
? PROGRESSION
It is not always possible to write each and every sequence of some rule.
For example : Sequence of prime numbers 2, 3, 5, 7, 11,... cannot be expressed explicitly by stating a rule
and we do not have any expression for writting the general term of this sequence.
The sequence that follows a certain pattern is called a progression. Thus, the sequence 2, 3, 5, 7, 11,.... is
not a progression. In a progression, we can always write the nth term.
Consider the following collection of numbers : (i) 1, 3, 5, 7...     (ii)
1 1 1 1
, , , ,...
2 3 4 5
From the above collection of numbers, we observe that
( i ) Each term is greater than the previous by 2.
( i i ) In each term the numerator is 1 and the denominator is obtained by adding 1 to the preceding denominator.
Thus, we observe that the collection of numbers given in (i) and (ii) follow a certain pattern and as such are
all progressions.
? ARITHMETIC PROGRESSIONS
An arithmetic progression is that list of numbers in which the first term is given and each term, other
than the first term is obtained by adding a fixed number 'd' to the preceding term.
The fixed term 'd' is known as the common difference of the arithmetic progression. It's value can be positive,
negative or zero. The first term is denoted by 'a' or 'a
1
' and the last term by ' ?'.
Ex. Consider a sequence 6, 10, 14, 18, 22, .....
Here, a
1
= 6, a
2
= 10, a
3
= 14, a
4
= 18, a
5
= 22
a
2
– a
1
= 10 – 6 = 4
a
3
– a
2
= 14 – 10 = 4
a
4
– a
3
= 18 – 14 = 4
- - - - - - - - - - - - - - - - -
- - - - - - - - - - - - - - - - -
Therefore, the sequence is an arithmetic progression in which the first term a = 6 and the common difference d = 4.
Symbolical form : Let us denote the first term of an AP by a
1
, second term by a
2
, .... nth term by a
n
and
the common difference by d. Then the AP becomes a
1
, a
2
, a
3
,....,a
n
.
So, a
2
– a
1
= a
3
– a
2
= ... = a
n
– a
n–1
= d.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
4
General form : In general form, an arithmetic progression with first term 'a' and common difference 'd' can
be represented as follows :
a, a + d, a + 2d, a + 3d, a + 4d,....
Finite AP : An AP in which there are only a finite number of terms is called a finite AP. It may
be noted that each such AP has a last term.
Ex. ( a ) The heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,
148, 149,..., 157.
( b ) The minimum temperatures (in degree Celsius) recorded for a week in the month of January in a city arranged
in ascending order are –3.1, – 3.0, –2.9, – 2.8, – 2.7, – 2.6, – 2.5
Infinite AP : An AP in which the number of terms is not finite is called infinite AP. It is note worthy
that such APs do not have a last term.
Ex. ( a ) 1, 2, 3, 4,......
( b ) 100, 70, 40, 10,.....
Least Information Required : To know about an AP, the minimum information we need to know is to know
both – the first term a and the common difference d.
For instance if the first term a is 6 and the common difference d is 3, then AP is 6, 9, 12, 15,...
Similarly, when
a = – 7, d = – 2, the AP is –7, –9, –11, –13,...
a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3,...
So if we know what a and d are we can list the AP.
Ex.4 In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
( i ) The taxi fare after each km  when the fare is Rs. 15 for the first km and Rs 8 for each additional km.
( i i ) The amount of air present in a cylinder when a vacuum pump removes
1
4
of the air remaining in the
cylinder at a time.
( i i i ) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and
rises by Rs. 50 for each subsequent metre.
( i v ) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at
8% per annum. [NCERT]
Sol. ( i ) Taxi fare for 1 km = Rs. 15 = a
1
Taxi fare for 2 kms = Rs. 15 + 8 = Rs. 23 = a
2
Taxi fare for 3 kms = Rs. 23 + 8 = Rs. 31 = a
3
Taxi fare for 4 kms = Rs. 31 + 8 = Rs. 39 = a
4
and so on.
a
2
– a
1
= Rs. 23 – 15 = Rs. 8
a
3
– a
2
= Rs. 31 – 23 = Rs. 8
a
4
– a
3
= Rs. 39 – 31 = Rs. 8
i.e., a
k+1
– a
k
is the same everytime.
So, this list of numbers form an arithmetic progression with the first term a = Rs 15 and the common
difference d = Rs. 8.
( i i ) Amount of air present in the cylinder = x units (say) = a
1
Amount of air present in the cylinder after one time removal of air by the vacuum pump = x –
x
4
=
3 x
4
units = a
2
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 4
Amount of air present in the cylinder after two times removal of air by the vacuum pump =
3 x
4
–
1
4
3 x
4
? ?
? ?
? ?
=
3 x
4
–
3 x
16
=
9 x
16
units =
2
3
4
? ?
? ?
? ?
x units = a
3
Amount of air present in the cylinder after three times removal of air by the vacuum pump
=
2
3
4
? ?
? ?
? ?
x –
1
4
2
3
4
? ?
? ?
? ?
x ? ?
1
1 –
4
? ?
? ?
? ?
2
3
4
? ?
? ?
? ?
x  ? ?
3
4
? ?
? ?
? ?
2
3
4
? ?
? ?
? ?
x =
3
3
4
? ?
? ?
? ?
x units = a
4
and so on.
a
2
– a
1
=
3 x
4
– x =
x
4
units
a
3
– a
2
=
2
3
4
? ?
? ?
? ?
x –
3
4
x =
3
–
16
x units
As a
2
– a
1
? a
3
– a
2
, this list of numbers does not form an AP.
( i i i ) Cost of digging the well after 1 metre of digging = Rs. 150 = a
1
Cost of digging the well after 2 metres of digging = Rs. 150 + 50 = Rs 200 = a
2
Cost of digging the well after 3 metres of digging = Rs. 200 + 50 = Rs 250 = a
3
Cost of digging the well after 4 metres of digging = Rs. 250 + 50 = Rs 300 = a
4
and so on.
a
2
– a
1
= Rs 200 – 150 = 50
a
3
– a
2
= Rs 250 – 200 = 50
a
4
– a
3
= Rs 300 – 250 = 50
i.e., a
k–1
– a
k
is the same everytime. So this list of numbers forms an AP with the first term a = Rs.
150 and the common difference d = Rs. 50
( i v ) Amount of money after 1 year = Rs. 10000
8
1
100
? ?
?
? ?
? ?
= a
1
Amount of money after 2 years = Rs. 10000
2
8
1
100
? ?
?
? ?
? ?
= a
2
Amount of money after 3 years = Rs. 10000
3
8
1
100
? ?
?
? ?
? ?
= a
3
Amount of money after 4 years = Rs. 10000
4
8
1
100
? ?
?
? ?
? ?
= a
4
a
2
– a
1
= Rs. 10000
2
8
1
100
? ?
?
? ?
? ?
– Rs. 10000
8
1
100
? ?
?
? ?
? ?
= Rs. 10000
8
1
100
? ?
?
? ?
? ?
8
1 – 1
100
? ?
?
? ?
? ?
? ?Rs. 10000
8
1
100
? ?
?
? ?
? ?
8
100
? ?
? ?
? ?
a
3
– a
2
= Rs. 10000
3
8
1
100
? ?
?
? ?
? ?
– Rs. 10000
2
8
1
100
? ?
?
? ?
? ?
? ?Rs. 10000
2
8
1
100
? ?
?
? ?
? ?
8
1 – 1
100
? ?
?
? ?
? ?
= Rs. 10000
2
8
1
100
? ?
?
? ?
? ?
8
100
? ?
? ?
? ?
As a
2
– a
1
? a
3
– a
2
, this list of numbers does not form an AP.
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;