Careers360 Logo
Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF

Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF

Edited By Safeer PP | Updated on Apr 07, 2025 09:41 AM IST

Arithmetic progression is a sequence of numbers where the numbers follow an order and have a constant difference between them. An arithmetic progression is also known as an arithmetic sequence, where the difference between the previous and next term is the same. In an arithmetic progression, the next term is generally determined by applying some arithmetic operations. An arithmetic progression is required in many fields, like physics, engineering, economics, etc., for predicting sequences or analysing trends by observing the rate of change.

This Story also Contains
  1. NCERT Class 10 Maths Chapter 5 Arithmetic Progressions: Notes
  2. Class 10 Chapter Wise Notes
  3. NCERT Class 10 Exemplar Solutions for Maths And Science
Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF
Arithmetic Progressions Class 10th Notes - Free NCERT Class 10 Maths Chapter 5 Notes - Download PDF

These notes covered all the topics and subtopics of arithmetic progression, like the basic definition of arithmetic progression, finite and infinite arithmetic progression, general term of arithmetic progression, sum of the nth term and arithmetic mean and formulae required to calculate the arithmetic progression. CBSE Class 10 chapter Arithmetic Progression also includes the nth term of the arithmetic progression. Students must practice all the topics of arithmetic progression and their examples from the NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression. Students must practice questions and check their solutions using the NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression.

NCERT Class 10 Maths Chapter 5 Arithmetic Progressions: Notes

Arithmetic Progression: Arithmetic progression is a sequence, list or series of numbers where the difference between two terms is constant, and we can get the preceding and succeeding terms by performing the arithmetic operation.
Arithmetic progression is denoted by AP, and the first term is denoted by a1 and the last term is denoted by an. It implies that the number of terms in AP is n.
Example: 5, 8, 11, 14, 17, 20, 23......

Common difference: In an arithmetic progression, the difference between two consecutive terms is called the common difference, and it is denoted by d.
Example: 7, 12, 17, 22, 27, 32
In this arithmetic progression, the common difference (d) = 5

Formula For Calculating Common Difference:
d = (an an - 1)
Where,
d = Common difference
an = nth term of the sequence
an - 1) = (n - 1)th term of the sequence
There are three types of common difference (d):

1. Positive Difference: The AP is increasing in this case.
Example: 1, 2, 3, 4, 5

2. Negative Difference: In this case, the AP decreases.
Example: 100, 90, 80, 70, 60, 50

3. Zero Difference: In this case, the AP is constant.
Example: 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Types of Arithmetic Progression (AP):

There are two types of arithmetic progression (AP):

1. Finite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is finite is called a finite arithmetic progression (AP). The finite AP have an nth term.
Example: 1, 3, 5, 7, 9, 11

2. Infinite Arithmetic Progression: An arithmetic progression (AP) in which the number of terms is infinite is called an infinite arithmetic progression (AP). The infinite AP does not have the nth term.
Example: 2, 4, 6, 8, 10, ........

The nth Term Of An Arithmetic Progression (AP):

The nth term in the last term in AP and the formula for determining the nth term is as follows:
an = a+(n1)d
Here,
a = First term
d = Common difference
n = number of terms

Example: Determine the 10th term of the arithmetic progression 4, 8, 12, 16, .......
Given:
AP = 4, 8, 12, 16, .......
Here, a = 4
(8 - 4) = 4 and (12 - 8) = 4
So, d = 4
And now, we have to find the 10th term of the given AP.
Therefore, n = 10
an = a+(n1)d
After substituting the values in the formula, we get
a10 = 4 + (10 – 1)4
a10 = 4 + 9 × 4
a10 = 4 + 36
a10 = 40
So, the 10th term of the given arithmetic sequence 4, 8, 12, 16, ....... is 40

The General Form Of An Arithmetic Progression (AP):

The general form of an AP is as follows:
a, a + d, a + 2d, a + 3d, a + 4d ....
Where a is the first term and d is the common difference, the value of d = 0, d > 0 or d < 0.

The Sum Of The Terms In Arithmetic Progression (AP):

The formula for calculating the sum of the terms in AP is as follows:
S = n2 [2a+(n1)d]
Where,
S = Sum of the terms
n = Number of terms
a = First term
d = Common difference
Or we can also write this as,
S = n2 (a+an)

Example: Determine the sum of the first 18 terms of AP 17, 11, 5, -1, .......
Given:
AP = 17, 11, 5, -1, .......
First term a = 17
Common difference d = 11 - 17 = -6, and 5 - 11 = -6
Therefore, the common difference d = -6
Number of terms n = 18
S = n2 [2a+(n1)d]
After substituting all the values, we get
S = 182 [2×17+(181)×(6)]
S = 9 × [34+(17)×(6)]
S = 9 × [34+(102)]
S = 9 × (68)
S = 612
Therefore, the sum of the 18th term is -612.

Arithmetic Mean (AM):

The average of the terms in an arithmetic progression is called the Arithmetic Mean (AM). Therefore, the arithmetic mean formula is as follows:
AM = Sum of the termsNumber of the terms

The Sum of the First n Natural Numbers:

The sum of the first n natural numbers can be determined by:
Sn = n(n+1)2
Here,
n = Number of terms

Example: Determine the sum of the first 15 natural numbers.
Given:
n = 15
Sn = n(n+1)2
Sn = 15(15+1)2
Sn = 15×162
Sn = 120
Therefore, the sum of the first 15 natural numbers is 120.

Class 10 Chapter Wise Notes

Students must download the notes below for each chapter to ace the topics.

NCERT Class 10 Solutions for Maths And Science
Students must check the NCERT solutions for Class 10 Maths and Science given below:

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

NCERT Class 10 Exemplar Solutions for Maths And Science

Students must check the NCERT exemplar solutions for Class 10 Maths and Science given below:

Read More: Syllabus and Books

Frequently Asked Questions (FAQs)

1. What is an arithmetic progression (AP) in Class 10 Maths?

An arithmetic progression is a sequence, list or series of numbers where the difference between two terms is constant, and we can get the preceding and succeeding terms by performing the arithmetic operation.

2. What are the important formulas of Arithmetic Progression?

The important formulas in Arithmetic progression are:
The common difference in an Arithmetic Progression (AP) = d = (an  an - 1)
The nth term of an Arithmetic Progression (AP) = a= a+(n1)d
The sum of the terms in Arithmetic Progression (AP) = Sn2 [2a+(n1)d]
Arithmetic Mean = AM = Sum  of  the  termsNumber of  the  terms
The sum of the n natural numbers = Sn = n(n+1)2

3. How do you find the nth term of an AP?

Students can determine that the nth term of the arithmetic progression is a= a+(n1)d, where a = First term, d = Common difference, n = number of terms.

4. What is the sum of n terms of an AP?

The formula for calculating the sum of the terms in AP is S = n2 [2a+(n1)d], Where, S = Sum of the terms, n = Number of terms, a = First term, d = Common difference

5. What is the common difference in an AP?

In an arithmetic progression, the difference between two consecutive terms is called the common difference, and it is denoted by d.

Articles

Upcoming School Exams

Application Date:24 March,2025 - 23 April,2025

Admit Card Date:25 March,2025 - 21 April,2025

Admit Card Date:25 March,2025 - 17 April,2025

View All School Exams

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top