NCERT Textbook - Arithmetic Progressions Class 10 Notes | EduRev

Class 10 : NCERT Textbook - Arithmetic Progressions Class 10 Notes | EduRev

Created by: Indu Gupta
 Page 1


ARITHMETIC PROGRESSIONS 93
5
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such
examples are :
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes 
5
4
 times of itself after every 3 years.
The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12
years will be, respectively :
10000, 12500, 15625, 19531.25
ARITHMETIC PROGRESSIONS
Fig. 5.1
Page 2


ARITHMETIC PROGRESSIONS 93
5
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such
examples are :
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes 
5
4
 times of itself after every 3 years.
The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12
years will be, respectively :
10000, 12500, 15625, 19531.25
ARITHMETIC PROGRESSIONS
Fig. 5.1
94 MATHEMA TICS
(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)
are, respectively
1
2
, 2
2
, 3
2
, . . . .
Fig. 5.2
(v) Shakila put Rs 100 into her daughter’s money box when she was one year old
and increased the amount by Rs 50 every year. The amounts of money (in Rs) in
the box on the 1st, 2nd, 3rd, 4th, . . . birthday were
100, 150, 200, 250, . . ., respectively.
(vi) A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the 1st, 2nd, 3rd, . . ., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig. 5.3
Page 3


ARITHMETIC PROGRESSIONS 93
5
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such
examples are :
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes 
5
4
 times of itself after every 3 years.
The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12
years will be, respectively :
10000, 12500, 15625, 19531.25
ARITHMETIC PROGRESSIONS
Fig. 5.1
94 MATHEMA TICS
(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)
are, respectively
1
2
, 2
2
, 3
2
, . . . .
Fig. 5.2
(v) Shakila put Rs 100 into her daughter’s money box when she was one year old
and increased the amount by Rs 50 every year. The amounts of money (in Rs) in
the box on the 1st, 2nd, 3rd, 4th, . . . birthday were
100, 150, 200, 250, . . ., respectively.
(vi) A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the 1st, 2nd, 3rd, . . ., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig. 5.3
ARITHMETIC PROGRESSIONS 95
In the examples above, we observe some patterns. In some, we find that the
succeeding terms are obtained by adding a fixed number, in other by multiplying
with a fixed number, in another we find that they are squares of consecutive
numbers, and so on.
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 nth terms and the sum of n consecutive terms, and use this knowledge in
solving some daily life problems.
5.2 Arithmetic Progressions
Consider the following lists of numbers :
(i) 1, 2, 3, 4, . . .
(ii) 100, 70, 40, 10, . . .
(iii) –3, –2, –1, 0, . . .
(iv) 3, 3, 3, 3, . . .
(v) –1.0, –1.5, –2.0, –2.5, . . .
Each of the numbers in the list is called a term.
Given a term, can you write the next term in each of the lists above? If so, how
will you write it? Perhaps by following a pattern or rule. Let us observe and write the
rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to  the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the
term preceding it.
In all the lists above, we see that successive terms are obtained by adding a fixed
number to the preceding terms. Such list of numbers is said to form an  Arithmetic
Progression ( AP ).
So, an arithmetic progression is a list of numbers in which each term  is
obtained by adding a fixed number to the preceding term except the first
term.
This fixed number is called the common difference of the AP. Remember that
it can be positive, negative or zero.
Page 4


ARITHMETIC PROGRESSIONS 93
5
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such
examples are :
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes 
5
4
 times of itself after every 3 years.
The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12
years will be, respectively :
10000, 12500, 15625, 19531.25
ARITHMETIC PROGRESSIONS
Fig. 5.1
94 MATHEMA TICS
(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)
are, respectively
1
2
, 2
2
, 3
2
, . . . .
Fig. 5.2
(v) Shakila put Rs 100 into her daughter’s money box when she was one year old
and increased the amount by Rs 50 every year. The amounts of money (in Rs) in
the box on the 1st, 2nd, 3rd, 4th, . . . birthday were
100, 150, 200, 250, . . ., respectively.
(vi) A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the 1st, 2nd, 3rd, . . ., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig. 5.3
ARITHMETIC PROGRESSIONS 95
In the examples above, we observe some patterns. In some, we find that the
succeeding terms are obtained by adding a fixed number, in other by multiplying
with a fixed number, in another we find that they are squares of consecutive
numbers, and so on.
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 nth terms and the sum of n consecutive terms, and use this knowledge in
solving some daily life problems.
5.2 Arithmetic Progressions
Consider the following lists of numbers :
(i) 1, 2, 3, 4, . . .
(ii) 100, 70, 40, 10, . . .
(iii) –3, –2, –1, 0, . . .
(iv) 3, 3, 3, 3, . . .
(v) –1.0, –1.5, –2.0, –2.5, . . .
Each of the numbers in the list is called a term.
Given a term, can you write the next term in each of the lists above? If so, how
will you write it? Perhaps by following a pattern or rule. Let us observe and write the
rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to  the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the
term preceding it.
In all the lists above, we see that successive terms are obtained by adding a fixed
number to the preceding terms. Such list of numbers is said to form an  Arithmetic
Progression ( AP ).
So, an arithmetic progression is a list of numbers in which each term  is
obtained by adding a fixed number to the preceding term except the first
term.
This fixed number is called the common difference of the AP. Remember that
it can be positive, negative or zero.
96 MATHEMA TICS
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.
Some more examples of AP are:
(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
(c) The balance money ( in Rs ) after paying 5 % of the total  loan of Rs 1000 every
month is 950, 900, 850, 800, . . ., 50.
(d) The cash prizes ( in Rs ) given by a school to the toppers of Classes I to XII are,
respectively, 200, 250, 300, 350, . . ., 750.
(e) The total savings (in Rs) after every month for 10 months when Rs 50 are saved
each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that
a, a + d, a + 2d, a + 3d, . . .
represents an arithmetic progression where a is the first term and d the common
difference. This is called the general form of an AP.
Note that in examples (a) to (e) above, there are only a finite number of terms.
Such an AP is called a finite AP . Also note that each of these Arithmetic Progressions
(APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs
and so they are called infinite Arithmetic Progressions. Such APs do not have a
last term.
Now, to know about an AP , what is the minimum information that you need? Is it
enough to know the first term? Or, is it enough to know only the common difference?
You will find that you will need 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
the AP is
 6, 9,12, 15, . . .
and if a is 6 and d is – 3, then the AP is
6, 3, 0, –3, . . .
Page 5


ARITHMETIC PROGRESSIONS 93
5
5.1 Introduction
You must have observed that in nature, many things follow a certain pattern, such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the
spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such
examples are :
(i) Reena applied for a job and got selected. She
has been offered a job with a starting monthly
salary of Rs 8000, with an annual increment of
Rs 500 in her salary. Her salary (in Rs) for the
1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease
uniformly by 2 cm from bottom to top
(see Fig. 5.1). The bottom rung is 45 cm in
length. The lengths (in cm) of the 1st, 2nd,
3rd, . . ., 8th rung from the bottom to the top
are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes 
5
4
 times of itself after every 3 years.
The maturity amount (in Rs) of an investment of Rs 8000 after 3, 6, 9 and 12
years will be, respectively :
10000, 12500, 15625, 19531.25
ARITHMETIC PROGRESSIONS
Fig. 5.1
94 MATHEMA TICS
(iv) The number of unit squares in squares with side 1, 2, 3, . . . units (see Fig. 5.2)
are, respectively
1
2
, 2
2
, 3
2
, . . . .
Fig. 5.2
(v) Shakila put Rs 100 into her daughter’s money box when she was one year old
and increased the amount by Rs 50 every year. The amounts of money (in Rs) in
the box on the 1st, 2nd, 3rd, 4th, . . . birthday were
100, 150, 200, 250, . . ., respectively.
(vi) A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits
produce a new pair in their second month and in every subsequent month (see
Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of
the 1st, 2nd, 3rd, . . ., 6th month, respectively are :
1, 1, 2, 3, 5, 8
Fig. 5.3
ARITHMETIC PROGRESSIONS 95
In the examples above, we observe some patterns. In some, we find that the
succeeding terms are obtained by adding a fixed number, in other by multiplying
with a fixed number, in another we find that they are squares of consecutive
numbers, and so on.
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 nth terms and the sum of n consecutive terms, and use this knowledge in
solving some daily life problems.
5.2 Arithmetic Progressions
Consider the following lists of numbers :
(i) 1, 2, 3, 4, . . .
(ii) 100, 70, 40, 10, . . .
(iii) –3, –2, –1, 0, . . .
(iv) 3, 3, 3, 3, . . .
(v) –1.0, –1.5, –2.0, –2.5, . . .
Each of the numbers in the list is called a term.
Given a term, can you write the next term in each of the lists above? If so, how
will you write it? Perhaps by following a pattern or rule. Let us observe and write the
rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to  the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding
(or subtracting) 0 to the term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the
term preceding it.
In all the lists above, we see that successive terms are obtained by adding a fixed
number to the preceding terms. Such list of numbers is said to form an  Arithmetic
Progression ( AP ).
So, an arithmetic progression is a list of numbers in which each term  is
obtained by adding a fixed number to the preceding term except the first
term.
This fixed number is called the common difference of the AP. Remember that
it can be positive, negative or zero.
96 MATHEMA TICS
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.
Some more examples of AP are:
(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
(c) The balance money ( in Rs ) after paying 5 % of the total  loan of Rs 1000 every
month is 950, 900, 850, 800, . . ., 50.
(d) The cash prizes ( in Rs ) given by a school to the toppers of Classes I to XII are,
respectively, 200, 250, 300, 350, . . ., 750.
(e) The total savings (in Rs) after every month for 10 months when Rs 50 are saved
each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that
a, a + d, a + 2d, a + 3d, . . .
represents an arithmetic progression where a is the first term and d the common
difference. This is called the general form of an AP.
Note that in examples (a) to (e) above, there are only a finite number of terms.
Such an AP is called a finite AP . Also note that each of these Arithmetic Progressions
(APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs
and so they are called infinite Arithmetic Progressions. Such APs do not have a
last term.
Now, to know about an AP , what is the minimum information that you need? Is it
enough to know the first term? Or, is it enough to know only the common difference?
You will find that you will need 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
the AP is
 6, 9,12, 15, . . .
and if a is 6 and d is – 3, then the AP is
6, 3, 0, –3, . . .
ARITHMETIC PROGRESSIONS 97
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, . . .
a =  0, d = 1
1
2
, the AP is  0,  1
1
2
, 3, 4
1
2
, 6, . . .
a = 2, d = 0, the AP is  2, 2, 2, 2, . . .
So, if you know what a and d are, you can list the AP. What about the other way
round? That is, if you are given a list of numbers can you say that it is an AP and then
find a and d? Since a is the first term, it can easily be written. We know that in an AP ,
every succeeding term is obtained by adding d to the preceding term. So, d found by
subtracting any term from its succeeding term, i.e., the term which immediately follows
it should be same for an AP .
For example, for the list of numbers :
6, 9, 12, 15, . . . ,
We have a
2 
– a
1
 = 9  – 6 = 3,
a
3
 – a
2
 = 12 – 9 = 3,
a
4
 – a
3
 = 15 – 12 = 3
Here the difference of any two consecutive terms in each case is 3. So, the
given list is an AP whose first term a is 6 and common difference d is 3.
For the list of numbers : 6, 3, 0, – 3, . . .,
a
2 
– a
1 
= 3 – 6 = – 3
a
3
 – a
2 
= 0 – 3 = – 3
a
4
 – a
3 
= –3 – 0  = –3
Similarly this is also an AP whose first term is 6 and the common difference
is –3.
In general, for an AP a
1
, a
2
, . . ., a
n
, we have
d = a
k + 1
 – a
k
where a
k + 1
 
 
and a
k 
 are the ( k + 1)th and the kth terms respectively.
To obtain d in a given AP, we need not find all of a
2
 – a
1
, a
3
 – a
2
, a
4
 – a
3
, . . . .
It is enough to find only one of them.
Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the
difference between any two consecutive terms is not the same. So, this is not an AP .
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