NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions - Free PDF Download
In Maths NCERT Solutions Class 10 Chapter 5, students will learn about the arithmetic progression. The NCERT Solutions for Class 10 Maths Chapter 5 PDF file, available for free, can help students to score good marks. Students can download this PDF file by visiting Vedantu. This file is prepared by the best academic experts in India. Every answer is written according to the guidelines set by CBSE. Further, every single step is taken to ensure that students can score good marks.


Glance of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions | Vedantu
An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference d where
.The first term of an AP is denoted by a and the nth term by
.You can find nth term in the AP using the formula:
The sum of the first n terms
in an AP is calculated using the formula:The common difference (d) can be positive, negative, or zero.
Positive D: The sequence is increasing. (e.g., 2, 5, 8, 11, ...)
Negative D: The sequence is decreasing. (e.g., 8, 5, 2, -1, ...)
Zero D: The sequence is constant. (e.g., 4, 4, 4, 4, ...)
The graph of an AP is a straight line. The slope of this line is equal to the common difference (d).
The average of the first and last term is equal to the middle term if there are an even number of terms.
This article contains chapter notes, formula and exercises link and important questions for chapter 5 - Arithmetic Progressions.
There are four exercises (49 fully solved questions) in class 10th maths chapter 5 Arithmetic Progressions.
Access Exercise Wise NCERT Solutions for Chapter 5 Maths Class 10
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
Exercises under NCERT Class 10 Maths Chapter 5 - Arithmetic Progressions
Exercise 5.1 - This exercise contains four problems, each with multiple parts. These problems aim to introduce students to the fundamental formulas of Arithmetic Progressions. These formulas include those for finding the first and last terms, calculating the sum of an A.P., and finding an unknown term of an A.P. using the common difference, among others.
Exercise 5.2 - This is the second exercise contains 20 problems. This exercise covers several problems that involve finding the terms of an A.P. by using formulas and substituting known values of the A.P. The exercise also includes numerous word problems to help students understand how to apply A.P. formulas.
Exercise 5.3 - It contains 20 problems related to arithmetic progressions. The problems in this exercise range from very easy to difficult word problems. To solve these problems, students are advised to have a thorough understanding of all the A.P. formulas provided in the chapter.
Exercise 5.4 - It is the final exercise and contains five problems. The first two problems require students to find the first negative term of a given A.P. or the first term of a given A.P. from the sum and product of two other known terms of the same A.P. The remaining problems are word problems that involve basic applications of volume, length, and other physical measurements.
Access NCERT Solutions for Maths Chapter 5 – Arithmetic Progression
Exercise 5.1
1. In Which of the Following Situations, Does the List of Numbers Involved Make As Arithmetic Progression and Why?
(i). The Taxi Fare After Each Km When the Fare is Rs
Ans: Given the fare of first km is Rs.
Taxi fare for
Taxi fare for
Taxi fare for
Similarly, Taxi fare for
Therefore, we can conclude that the above list forms an A.P with common difference of
(ii). The Amount of Air Present in a Cylinder When a Vacuum Pump Removes a Quarter of the Air Remaining in the Cylinder at a Time.
Ans: Let the initial volume of air in a cylinder be
Volume after
Volume after
Volume after
Similarly, Volume after
We can observe that the subsequent terms are not added with a constant digit but are being multiplied by
(iii). The cost of digging a well after every meter of digging, when it costs Rs
Ans: Given the cost of digging for the first meter is Rs.
Cost of digging for
Cost of digging for
Cost of digging for
Similarly, Cost of digging for
Therefore, we can conclude that the above list forms an A.P with common difference of
(iv). The amount of money in the account every year, when Rs
Ans: Given the principal amount is Rs.
Amount after
Amount after
Amount after
Similarly, Amount after
We can observe that the subsequent terms are not added with a constant digit but are being multiplied by
2. Write first four terms of the A.P. when the first term
1.
Ans: We know that the
Substituting
Therefore, from (2)
2.
Ans: We know that the
Substituting
Therefore, from (2)
3.
Ans: We know that the
Substituting
Therefore, from (2)
4.
Ans: We know that the
Substituting
Therefore, from (2)
5.
Ans: We know that the
Substituting
Therefore, from (2)
3. For the following A.P.s, write the first term and the common difference.
1.
Ans: From the given AP, we can see that the first term is
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference
2.
Ans: From the given AP, we can see that the first term is
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference
3.
Ans: From the given AP, we can see that the first term is
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference
4.
Ans: From the given AP, we can see that the first term is
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference
4. Which of the following are AP’s? If they form an AP, find the common difference
1.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.
2.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
3.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
4.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
5.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
6.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.
7.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
8.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting
Therefore, from (5)
9.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
10.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
We know that the
Substituting,
we get,
11.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.
12.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term
Substituting, a =
we get,
13.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.
14.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.
15.
Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series form an A.P. with first term a and common difference is given by
Substituting a =
we get,
Exercise 5.2
1. Fill in the blanks in the following table, given that
I | ||||
II | ||||
III | ||||
IV | ||||
V |
(i). Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
(ii). Ans: Given, the first Term,
Given, the
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
(iii). Ans: Given, the
Given, the common Difference,
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
(iv) Ans: Given, the first Term,
Given, the common Difference,
Given, the
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
(v). Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
2. Choose the correct choice in the following and justify
(i).
Ans: C.
Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
(ii).
Ans: B.
Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
3. In the following APs find the missing term in the blanks
(i).
Ans: Given, first term
We know that the
Substituting the values from (1) we get,
Given, third term
From (1), (2) and (3) we get for
(ii).
Ans: Given, second term
We know that the
Substituting the values from (1) for
Given, fourth term
Solving (3) and (4) by subtracting (3) from (4) we get,
From (3) and (5) we get
Substituting the values from (5) and (6) in (2) we get,
First term,
(iii).
Ans: Given, first term
We know that the
Substituting the values from (1) in (2) we get,
Given, fourth term
From (3) and (4) we get
Second term,
(iv).
Ans: Given, first term
We know that the
Substituting the values from (1) in (2) we get,
Given, sixth term
From (3) and (4) we get
Second term
(v).
Ans: Given, second term
We know that the
Substituting the values from (1) for
Given, sixth term
Solving (3) and (4) by subtracting (3) from (4) we get,
From (3) and (5) we get
Substituting the values from (5) and (6) in (2) we get,
First term,
4. Which term of the A.P.
Ans: Given, the first Term,
Given, the common Difference,
Given, the
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
Therefore,
5. Find the number of terms in each of the following A.P.
(i).
Ans: Given, the first Term,
Given, the common Difference,
Given, the
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
Therefore, given A.P. series has
(ii).
Ans: Given, the first Term,
Given, the common Difference,
Given, the
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
Therefore, given A.P. series has
6. Check whether
Ans: Given, the first Term,
Given, the common Difference,
Given, the
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
Since
7. Find the
Ans: Given, the
Given, the
We know that the
Substituting the values from (1) in (3) we get,
Substituting the values from (2) in (3) we get,
Solving equations (4) and (5) by subtracting (4) from (5) we get,
Substituting value from (6) in (4) we get,
Again, substituting the values from (6) and (7) in (3) we get,
To find the
Therefore, the
8. An A.P. consists of
Ans: Given, the
Given, the
We know that the
Substituting the values from (1) in (3) we get,
Substituting the values from (2) in (3) we get,
Solving equations (4) and (5) by subtracting (4) from (5) we get,
Substituting value from (6) in (4) we get,
Again, substituting the values from (6) and (7) in (3) we get,
To find the
Therefore, the
9. If the
Ans: Given, the
Given, the
We know that the
Substituting the values from (1) in (3) we get,
Substituting the values from (2) in (3) we get,
Solving equations (4) and (5) by subtracting (4) from (5) we get,
Substituting value from (6) in (4) we get,
Again, substituting the values from (6) and (7) in (3) we get,
T find the term which is zero, substitute
Therefore, given A.P. series has
10. If
Ans: Given that the
We know that the
For
For
Therefore, from (1), (3) and (4) we get,
Therefore, the common difference is
11. Which term of the A.P.
Ans: Let
We know that the
For
Therefore, from (1), (2) and (3) we get,
Now, given A.P.
Common difference
Hence, from (4) and (5) we get
Therefore,
12. Two APs have the same common difference. The difference between their
Ans: Let
(Since common difference is same)
Given that the difference between their
We know that the
Therefore, from (3) and (4) we get,
Similarly, the difference between their
Therefore, the difference between their
13. How many three-digit numbers are divisible by
Ans: First three-digit number that is divisible by
Therefore, the series becomes
This is an A.P. having first term as
Now, the largest
Leu us divide it by
Therefore,
Also, this will be the last term of the A.P. series.
Hence the final series is as follows:
Let 994 be the
Then,
Therefore,
14. How many multiples of
Ans: First number that is divisible by
Therefore, the series becomes
This is an A.P. having first term as
Now, the largest number in range is
Leu us divide it by
Therefore,
Hence the final series is as follows:
Let
Then,
Therefore,
15. For what value of
Ans: Given
Its first term is
Its first term is
We know that the
Therefore, from (1) and (3) we get the
And from (2) and (3) we get the
If the
Therefore, the
16. Determine the A.P. whose third term is
Ans: Given the
We know that the
For
For
Therefore, from (1), (3) and (4) we get,
Substituting (5) in (2) we get,
Given the third term of the A.P. is
Hence from (6),
Therefore, the A.P. will be
17. Find the
Ans: Given A.P.
Required A.P. is
Its first A.P. is
We know that the
Hence from (2) and (3) we get,
Substitute
Therefore,
18. The sum of
Ans: Given the sum of
Given the sum of
We know that the
For
For
For
For
Therefore, from (1), (6) and (4) we get,
From (2), (5) and (7) we get,
Subtracting (8) from (9) we get,
Substituting this value from (10) in (9) we get,
Therefore from (10) and (11), the first three terms of the A.P. are
19. Subba Rao started work in
Ans: Given in the first year, annual salary is Rs
In the second year, annual salary is Rs
In the third year, annual salary is Rs
This series will form an A.P. with first term
We know that the
Therefore, In the
To find the year in which his annual income reaches Rs
Therefore, in
20. Ramkali saved Rs
Ans: Given in the first week the savings is Rs
In the second week the savings is Rs
In the third week the savings is Rs
This series will form an A.P. with first term
We know that the
Therefore, In the
To find the week in which her savings reaches Rs
Therefore, in
Exercise 5.3
1. Find the sum of the following APs.
(i).
Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the sum of
Substituting the values from (1), (2) and (3) in (4) we get,
(ii).
Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the sum of
Substituting the values from (1), (2) and (3) in (4) we get,
(iii).
Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the sum of
Substituting the values from (1), (2) and (3) in (4) we get,
(iv).
Ans: Given, the first Term,
Given, the common Difference,
Given, the number of Terms,
We know that the sum of
Substituting the values from (1), (2) and (3) in (4) we get,
2. Find the sums given below
(i).
Ans: Given, the first Term,
Given, the common Difference,
We know that the
Substituting the values from (1) and (2) in (3) we get,
Given, last term of the series,
Substituting (5) in (4) we get,
We know that the sum of
Substituting the values from (1), (5) and (6) in (7) we get,
(ii).
Ans: Given, the first Term,
Given, the common Difference,
We know that the
Substituting the values from (1) and (2) in (3) we get,
Given, last term of the series,
Substituting (5) in (4) we get,
We know that the sum of
Substituting the values from (1), (5) and (6) in (7) we get,
(iii).
Ans: Given, the first Term,
Given, the common Difference,
We know that the
Substituting the values from (1) and (2) in (3) we get,
Given, last term of the series,
Substituting (5) in (4) we get,
We know that the sum of
Substituting the values from (1), (5) and (6) in (7) we get,
3. In an AP
(i). Given
Ans: Given, the first Term,
Given, the common Difference,
Given,
We know that the
Substituting the values from (1), (2) and (3) in (4) we get,
Simplifying it further we get,
We know that the sum of
Substituting the values from (1), (2) and (5) in (6) we get,
(ii). Given
Ans: Given, the first Term,
Given,
We know that the
Substituting the values from (1), (2) in (3) we get,
Simplifying it further we get,
We know that the sum of
Substituting the values from (1) and (4) in (5) we get,
(iii). Given
Ans: Given, the common difference,
Given,
We know that the
Substituting the values from (1), (2) in (3) we get,
Simplifying it further we get,
We know that the sum of
Substituting the values from (1) and (4) in (5) we get,
(iv). Given
Ans: Given,
Given, the sum of terms,
We know that the
Substituting the values from (1) in (3) we get,
We know that the sum of
Substituting the values from (1) in (5) we get,
Let us solve equations (4) and (5) by subtracting twice of (4) from (5) we get,
From (4) and (6) we get,
From (3), (6) and (7) for
(v). Given
Ans: Given, common difference,
Given, the sum of terms,
We know that the sum of
Substituting the values from (1), (2) in (3) we get,
We know that the
Substituting the values from (1), (4) in (5) we get,
(vi) Given
Ans: Given, common difference,
Given, first term,
Given, the sum of terms,
We know that the sum of
Substituting the values from (1), (2), (3) in (4) we get,
We know that the
Substituting the values from (1), (2), (4) in (5) we get,
(vii). Given
Ans: Given, first term,
Given, the sum of terms,
Given, the
We know that the sum of
Substituting the values from (1), (2) in (4) we get,
We know that the
Substituting the values from (1), (3) in (5) we get,
Let us solve equations (4) and (6) by subtracting
Substituting the values from (7) in (6) we get,
(viii). Given
Ans: Given, common difference,
Given, the sum of terms,
Given, the
We know that the sum of
Substituting the values from (1), (2) in (4) we get,
We know that the
Substituting the values from (1), (3) in (6) we get,
Let us solve equations (5) and (7) by substituting the value of
Substituting the values from (8) in (7) we get,
(ix). Given
Ans: Given, first term,
Given, the sum of terms,
Given, the number of terms,
We know that the sum of
Substituting the values from (1), (2) in (4) we get,
(x). Given
Ans: Given, last term,
Given, the sum of terms,
Given, the number of terms,
We know that the sum of
Substituting the values from (1), (2) in (4) we get,
4. How many terms of the A.P.
Ans: Given, common difference,
Given, first term,
Given, the sum of terms,
We know that the sum of
Substituting the values from (1), (2), (3) in (4) we get,
Since
5. The first term of an AP is
Ans: Given, first term,
Given, the sum of terms,
Given, the
We know that the sum of
Substituting the values from (1), (2), (3) in (4) we get,
We know that the
Substituting the values from (1), (3) in (5) we get,
6. The first and the last term of an AP are
Ans: Given, first term,
Given, the common difference,
Given, the
We know that the
Substituting the values from (1), (2), (3) in (4) we get,
We know that the sum of
Substituting the values from (1), (5), (3) in (6) we get,
7. Find the sum of first
Ans: Given, the common difference,
Given, the
We know that the
Substituting the values from (1), (2) in (3) we get,
We know that the sum of
Substituting the values from (1), (2), (4) in (5) we get,
8. Find the sum of first
Ans: Given, the
Given, the
We know that the
Substituting the values from (1) in (3) we get,
Substituting the values from (2) in (3) we get,
Solving equations (4) and (5) by subtracting (4) from (5) we get,
Substituting the value from (6) in (4) we get
We know that the sum of
Substituting the values from (7), (6) in (8) we get for
9. If the sum of first
Ans: Given, the sum of first
Given, the sum of first
We know that the sum of
Substituting the values from (1) in (3) we get,
Substituting the values from (2) in (3) we get,
Solving equations (4) and (5) by subtracting (4) from (5) we get,
Substituting the value from (6) in (4) we get
Substituting the values from (7), (6) in (3) we get,
10. Show that
(i).
Ans: Consider two consecutive terms of the given sequence. Say
Which is a constant
For
Therefore, it is an A.P. with first term
We know that the sum of
Therefore,
(ii).
Ans: Consider two consecutive terms of the given sequence. Say
Which is a constant
For
Therefore, it is an A.P. with first term
We know that the sum of
Therefore,
11. If the sum of the first
Ans: Given, the sum of the first
First term
Sum of first two terms
From (1) and (2),
Sum of first three terms
From (3) and (2),
Similarly,
Sum of first
Sum of first
From (4) and (5),
From (6),
12. Find the sum of first
Ans: First positive integer that is divisible by
Second positive integer that is divisible by
Third positive integer that is divisible by
Hence, it is an A.P. with first term and common difference both as
We know that the sum of
Therefore, for
13. Find the sum of first
Ans: First positive integer that is divisible by
Second positive integer that is divisible by
Third positive integer that is divisible by
Hence, it is an A.P. with first term and common difference both as
We know that the sum of
Therefore, for
14. Find the sum of the odd numbers between
Ans: The odd numbers between
It is an A.P. with first term
We know that the
Substitute
We know that the sum of
Substituting values from (1), (3) in (4) we get,
15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs.
Ans: Penalty of delay for first day is Rs.
Penalty of delay for second day is Rs.
Penalty of delay for third day is Rs.
Hence it is an A.P. with first term
Money the contractor has to pay as penalty, if he has delayed the work by
We know that the sum of
Therefore, the contractor has to pay Rs
16. A sum of Rs
Ans: Let the first prize be of Rs.
Therefore, it is an A.P. with first term
Given,
We know that the sum of
Therefore, the value of each of the prizes was
17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant
Ans: Each section of class I will plant
Each section of class II will plant
Each section of class III will plant
Therefore, it is an A.P series with first term and common difference both as
We know that the sum of
Therefore,
18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A of radii

Ans: Length of first semi-circle
Length of second semi-circle
Length of third semi-circle
Therefore, it is an A.P series with first term and common difference both as
We know that the sum of
Therefore, the length of such spiral of thirteen consecutive semi-circles
will be
19. The

Ans: Total logs in first row are
Total logs in second row are
Total logs in third row are
Therefore, it is an A.P series with first term
We know that the sum of
For
Total logs in
Therefore,
20. In a potato race, a bucket is placed at the starting point, which is
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
(Hint: to pick up the first potato and the second potato, the total
distance (in metres) run by a competitor is

Ans: Total distance run by competitor to collect and drop first potato
Total distance run by competitor to collect and drop second potato
Total distance run by competitor to collect and drop third potato
Therefore, it is an A.P series with first term
We know that the sum of
Therefore, the competitor will run a total distance of
Exercise 5.4
1. Which term of the A.P.
(Hint: Find
Ans: Given A.P.
Its first term is
We know that the
Therefore the
To find negative term, find
Hence from (1),
Therefore, the
2. The sum of the third and the seventh terms of an A.P is
Ans: Given the sum of third and seventh term of A.P.,
Given the sum of third and seventh term of A.P.,
We know that the
For
For
From (1),
From (2),
Let us now solve equations (3) and (4) by substituting the value of
CASE 1: For
Substitute
Therefore, it is an A.P series with first term
We know that the sum of
CASE 2: For
Substitute
Therefore, it is an A.P series with first term
3. A ladder has rungs
(Hint: number of rungs

Ans: Distance between first and last rungs is
Distance between two consecutive rungs is
Therefore, total number of rungs are
Also, we can observe that the length of each rung is decreasing in a uniform order. So, we can conclude that the length of rungs is in A.P. with first term
We know that the sum of
Therefore, the length of the wood required for the rungs is
4. The houses of a row are number consecutively from
Find this value of
(Hint:
Ans: Given houses are numbered
Clearly, they are numbered in A.P. series with both first term and common difference as
Now, there is house numbered

We know that the sum of
Therefore, house number
5. A small terrace at a football ground comprises of

Ans: Given that a football ground comprises of

Here blue step is the lowermost step. Let it be known as step
The red step is the second lowermost step. Let it be known as step
The green step is the third lower step. Let it be known as step

We can see that the height is increasing with each increasing step by a factor of
Therefore, we can conclude that the volume of steps is in A.P. with first term and common difference both as
We know that the sum of
Therefore, volume of concrete required to build the terrace is
NCERT Solutions for Class 10 Maths Chapter 5 solutions PDF download
5.1 Introduction: In this chapter, we'll understand the unique number patterns and will learn how to identify them, finding specific numbers in the given sequence and also how to find the sum of all numbers in a given sequence of numbers.
5.2 Arithmetic Progression: In simple terms, Arithmetic Progression or AP can be defined as a sequence of numbers. This sequence exists in an order in which the difference between any two consecutive numbers would be constant.
For Example: if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is a series of natural numbers. It can be said that this series is also an arithmetic progression between the difference between every two successive terms is 1. If we have a similar series of odd and even numbers, still, the difference between two successive terms will be two. This means that the odd and even number series will also be arithmetic progressions.
There are three different types of progressions. These different types of progressions that are mentioned in Chapter 5 Class 10 Maths NCERT book are:
Arithmetic Progression (AP)
Geometric Progression (GP)
Harmonic Progression (HP)
There are also three other definitions of arithmetic progression that students should remember for writing Arithmetic Progression Class 10 NCERT Solutions. These definitions are:
Definition One: Arithmetic progression is a mathematical sequence in which the difference between any two consecutive terms is always a constant. It can also be abbreviated as AP
Definition Two: It is also mentioned in NCERT Solutions Class 10 Maths Chapter 5 that arithmetic progression or sequence is the sequence of numbers. In this sequence of consecutive numbers, it is possible to find the next number by adding a fixed number to the previous number in the chain
Definition Three: In Arithmetic Progression Class 10 Solutions, the common difference of the AP is the fixed number that one should add to any term of the arithmetic progression. For example, in the arithmetic progression 1, 4, 7, 10, 13, 16, 19, 22, the value of the common difference is 3
In Class 10 Maths Chapter 5 Solutions, there are three main terms. These terms are:
The common difference (d)
nth Term (an)
The num of the first n terms (Sn)
These three terms are used to represent the property of arithmetic progression. In the next section, we will look at these three properties in more detail.
The Common Difference in Arithmetic Progression
For any given series of arithmetic progression, the terms that are used are the first term, the common difference between any two terms, and the nth term. Let’s assume that a1, a2, a3, a4, …, an is an arithmetic progression.
This means that the value of the common difference ‘d’ is:
Here, d is the value of the common difference. The value of d can be positive, negative, or zero.
The First Term of Arithmetic Progression
If an individual wants to write the arithmetic progression in terms of its common difference for solving, then it can be written as:
A, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …, a + (n - 1) d
In this sequence, a is the first term of the progression.
The General Form of an Arithmetic Progression
In this section, students will be able to do just that. Before we proceed, a student should begin with an assumption that the arithmetic progression for class 10 maths chapter 5 solutions is
Position of Terms | Representation of Terms | Values of Terms |
1 | A = a + (1 - 1) d | |
2 | A + d = a + (2 - 1) d | |
3 | A + 2d = a + (3 - 1) d | |
4 | A + 3d = a + (4 - 1) d | |
. | . | . |
. | . | . |
n | A + (n - 1) d |
5.3 The nth Term of Arithmetic Progression (AP)
This formula can be used for finding the class 10 maths chapter 5 NCERT solutions in which one needs to get the value of the nth term of an arithmetic progression. The formula can be written as:
Here, a is the first term, d is the value of the common difference, n is the number of terms, and an is the nth term.
Let’s take an example. Try to find out the nth term of the following arithmetic progression 1, 2, 3, 4, 5, …, an. The total number of terms is 15.
We know that n = 15. This means that according to the formula, we can say that:
Since, a = 1, and the common difference or d = 2 - 1 = 1
Then,
The behaviour of the entire sequence will also depend on the values of the common difference. This means that if the value of the common difference is positive, then the member terms will grow towards positive infinity. And if the value of the common difference is negative, then the member terms will move towards negative infinity.
5.4 The Sum of the First n Terms of an Arithmetic Progression(AP)
One can easily calculate the sum of n terms of any known progression. For an arithmetic progression, it is possible to calculate the sum of the first n terms if the value of the first term and the total terms are known. The formula is mentioned below.
But what if the value of the last term of the arithmetic progression is given? In that case, students should use the formula that is mentioned below.
S =
For ease of revision, we have also summarized all the major formulas of this chapter in a table. That table is mentioned below.
General Form of AP | A, a + d, a + 2d, a + 3d, a + 4d, …, a + nd |
The nth term of AP | |
Sum of n terms in AP | |
Sum of all terms in a finite AP with the last term as I |
Class 10 Maths Chapter 5: Exercises Breakdown
Chapter 5 -Arithmetic Progressions All Exercises in PDF Format | |
Exercise 5.1 | 4 Questions & Solutions |
Exercise 5.2 | 20 Questions & Solutions |
Exercise 5.3 | 20 Questions & Solutions |
Exercise 5.4 | 5 Questions & Solutions |
Conclusion
NCERT Solutions for Class 10 Maths Chapter on Arithmetic Progressions provided by Vedantu offer a comprehensive understanding of this fundamental concept.Students can gain a solid understanding of arithmetic progressions by concentrating on important concepts such as calculating the sum of terms, nth term, and common difference. It's important to pay close attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques.Arithmetic progressions are fundamental to more advanced mathematical ideas, hence it's important to understand them. This chapter has usually been the subject of five or six questions on previous year's exam papers. Therefore, practicing a variety of problems from NCERT Solutions and previous year papers can enhance preparation and confidence for exams.
Other Related Links for CBSE Class 10 Maths Chapter 5
S.No. | Important Links for Chapter 5 Arithmetic Progressions |
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Chapter-Specific NCERT Solutions for Class 10 Maths
Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
S.No. | NCERT Solutions Class 10 Chapter-wise Maths PDF |
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NCERT Study Resources for Class 10 Maths
For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.
FAQs on NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
1. I’m having doubts in Chapter 5 Maths NCERT. How can Vedantu help me?
If a student has any doubts, then he or she can always contact our in-house academic experts. We at Vedantu offer 24x7 query resolution services. Students can post their questions on the platform and one of our in-house experts will reply with the correct answer as soon as possible. Students can also take online classes. If there is any other way in which way we can help, you should contact us as soon as possible.
2. How can one find the sum of an arithmetic progression?
If one wants to find the sum of an arithmetic progression, then he or she must know the value of the first term, the number of terms, and the common difference that exists between each term. The following formula can be used to arrive at the final answer.
S = n / 2 (2 a + (n - 1) x d)
3. Mention the different types of progressions in mathematics.
There are three types of progressions in mathematics. These types of progressions are:
- Arithmetic progression (AP)
- Geometric progression (GP)
- Harmonic progression (HP)
4. Mention some uses of arithmetic progressions.
Arithmetic progressions can be used to generalize a set of patterns that we usually find in our daily lives.
5. What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5?
Vedantu offers solutions for the NCERT Textbook chapter-wise in a simplified manner. Chapter 5 of Class 10 Maths is Arithmetic Progressions. This chapter includes the sequence of numbers where the difference between the two consecutive numbers is the same. This chapter also finds applications in real life such as days of the week, students’ roll numbers, etc. This chapter also includes the other two types of progressions: Geometric and Harmonic.
6. Is it necessary to learn all the topics provided in NCERT Solutions for Class 10 Maths Chapter 5?
Yes, it is very important to learn all the topics and concepts provided in NCERT Solutions for Class 10 Maths Chapter 5. As this chapter forms the foundation for higher grade Mathematics, students need to know the basic formulae and their application. Vedantu offers solutions to all the exercises of chapter 5. The students can refer them online or download them for free to refer offline from the Vedantu website or the Vedantu app.
7. How many Exercises are there in Class 10 Maths Chapter 5 AP?
Chapter 5 of Class 10 Mathematics is Arithmetic progressions. This chapter includes four exercises with similar examples. The examples give you an idea to solve the exercise problems. Vedantu provides solutions to all the exercises in a step-by-step manner. One of the advantages of referring to Vedantu Solutions is that the answers are verified by the subject-matter-experts and are given in a simplified manner.
8. Is Chapter 5 Arithmetic Progression of Class 10 Maths interesting?
Yes, Chapter 5 Arithmetic Progression of Class 10 Maths is interesting. The concepts involved in this chapter are a foundation of the concepts that appear in the higher grades. Also, these have applications in real life such as roll numbers of the students in a class, months in a year, and weeks in a day. Vedantu provides the best learning while solving problems. Not only are the solutions verified, but they are also given in a simplified step-by-step manner.
9. Is Class 10 Arithmetic Progressions of Class 10 Hard?
Maths can be a tricky subject and as the board examinations approach, most of the students have fear while giving the examinations. The only solution to this is by achieving confidence and by practising different types of problems on the same concept. Vedantu offers the previous year’s question papers that the students can refer to get an idea of the marking scheme, the difficulty level of the question paper, and important questions.
10. Is arithmetic progression important?
Yes, arithmetic progression (AP) is important because it's a fundamental concept in mathematics with wide-ranging applications in various fields.
11. What is the main purpose of arithmetic?
The main purpose of arithmetic is to help us to perform fundamental mathematical operations, such as addition, subtraction, multiplication, and division. These functions are necessary for handling daily issues, handling money, calculating amounts, and recognising trends in numerical data. Whether it's counting objects, calculating the cost of groceries, or solving complex mathematical problems, arithmetic forms the basis of mathematical understanding and problem-solving skills.
12. Who is the father of arithmetic?
Brahmagupta is known as the father of arithmetic.
13. What is the full concept of arithmetic?
The full concept of arithmetic encompasses the study of basic mathematical operations involving numbers, such as addition, subtraction, multiplication, and division. It involves understanding how these operations work and how they can be applied to solve various mathematical problems. Arithmetic is fundamental to everyday life, from counting objects to managing finances and beyond. It forms the foundation of more advanced branches of mathematics and is essential for developing strong mathematical skills. In essence, arithmetic provides the tools and techniques necessary for performing calculations and solving numerical problems in a wide range of contexts.
14. What are the real life applications of arithmetic series?
Examples of Real-Life Arithmetic Sequences
Financial Planning
Population Growth
Seating around tables
Stacking cups, chairs, bowls











