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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 Mar 19, 2022 10:53 AM IST

NCERT Class 10 Maths chapter 5 notes based on arithmetic progression, in which general form of A.P and common difference are included. The chapter arithmetic progression Class 10 notes also include the arithmetic mean and the sum of n terms of an A.P.

The NCERT Class 10 chapter 1 notes include important formulas. CBSE Class 10 Maths chapter 1 notes contain systematic explanations of topics using examples and exercises. The important terms and techniques related to A.P. have also been included in CBSE Class 10 Maths chapter 1 notes. All these topics can be easily available and are downloaded from Class 10 Maths chapter 1 notes pdf download.

Also, students can refer,

Arithmetic Progression Class 10 Notes- Topic 1:

Sequence: This is the ordered list of numbers.

Arithmetic progression: The list of numbers is made in such a way that each term is having a common difference in between them and the comment difference is constant, that is adding a fixed number will proceed to the next term.

The General Form of A.P-

Lets suppose a is the first term and d is the common difference in between the terms.

Now, here a1=a

a2=a1+d= a + d

a2=a2+d= a + d + d = a + 2d

a3=a3+d= a + 2d + d = a + 3d



Moving further in the same way for “n” terms.

an=a +(n-1) d

Which is the nth term of an A.P. where;

a is the first term of an A.P.

n will be the number of terms and;

d is a common difference.

Sum of First n Terms of an A. P-

Now here in this scenario to get the value; considering the sum of first n terms of an A. P. is Sn.

Where,we can write the sum as follows:

Sn= (a) first term + (a+ d) second term + (a + 2d) third term +( a + 3d) third term………(1)

By putting the value of the product of ‘terms and the common difference we can write the series as:

Sn= (a +(n-1) d) +( a +(n-2) d) +(a +(n-3) d)+….+…. ………….(2)

Adding both the equation we get;

2Sn= n [2a +(n-1) d]

Sn=0.5n[2a +(n-1) d];

Where l = [a +(n-1) d]

So we can write the above equation is as follows.


Points to remember:

For calculating the Nth term of an AP=Sn-Sn-1

The sum of n positive integer will be calculated as Sn=n(n+1)/2

For n odd positive integers sum will be calculated as n2

For n even positive integers sum will be calculated as n(n+1)

Properties of Arithmetic Progression:

  • It should be noted that if in case constant is added or subtracted or multiplied to each and every term in an AP then we get the sequence which is also followed up with a new AP.

  • It should also be kept in mind that when each and every term of an AP is divided by a non-zero quantity then we get the result of a sequence of new AP.

Significance of NCERT Class 10 Maths Chapter 5 Notes

Arithmetic Progression Class 10 Notes will help to understand the formulas, statements, rules in detail. Also, this NCERT Class 10 Maths chapter 5 also contains previous year’s questions and NCERT TextBook pdf. It also contains FAQs or frequently asked questions along with topic-wise explanations. Class 10 Maths chapter 5 notes help to understand the concept of Sequence of Arithmetic progression as well as the general formula related to it.

Class 10 Chapter Wise Notes

NCERT Solutions of Cass 10 Subject Wise

NCERT Class 10 Exemplar Solutions for Other Subjects:

Frequently Asked Question (FAQs)

1. Using concepts in the NCERT Class 10 Maths chapter 5 notes solve for how- many two digits number will exactly be divisible by 6?

Let the sequence is taken as 12, 18, 24, …..96.

Here, a=12 , d=18-12=6, and last term  an=96


 nth term=a +(n-1) d


12+(n-1)6 = 96







2. Write the sum of first 10 natural number.


3. Write the sum of 1+3+5+7+9+11+13

The given sequence 1, 3, 5... is an ap



Get answers from students and experts

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)


Option 2)

\; K\;

Option 3)


Option 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)


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)


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

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