CBSE Class 10 Mathematics Chapter 5: Arithmetic Progressions NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 10 PDF

This comprehensive set of NCERT Solutions for CBSE Class 10 Mathematics, Chapter 5, focuses on Arithmetic Progressions (AP). Students will find detailed explanations and step-by-step solutions for identifying whether a given sequence of numbers forms an arithmetic progression. The solutions cover various real-world scenarios and mathematical problems, helping students understand the core concepts of AP, including the first term (a) and common difference (d). This resource is designed to reinforce learning, clarify doubts, and provide a solid foundation for exam preparation, enabling students to confidently tackle problems related to arithmetic progressions.

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Board	CBSE
Class	Class 10
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 5

Chapter summary

Chapter 5, Arithmetic Progressions, introduces students to sequences where the difference between consecutive terms is constant. This NCERT Solutions set provides clear explanations for identifying APs in different contexts, such as taxi fares or digging costs. It also guides students on how to generate the first few terms of an AP given the first term and common difference. The solutions focus on understanding the fundamental properties of APs, crucial for solving related problems in mathematics.

Learning outcomes

Understand the definition and properties of an Arithmetic Progression (AP).
Identify whether a given sequence of numbers forms an AP.
Determine the common difference between consecutive terms in an AP.
Calculate the first few terms of an AP given the first term and common difference.
Apply the concept of AP to real-world scenarios.

Topics covered

Paper topics

Introduction to Arithmetic Progressions (AP)
Identifying APs in real-world scenarios
Calculating the common difference
Determining the first few terms of an AP
Distinguishing AP from other sequences
Compound interest vs. simple interest progression

Important topics

Definition of Arithmetic Progression
Condition for a sequence to be an AP
Calculating the common difference (d)
Generating terms of an AP using 'a' and 'd'
Real-world applications of AP

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Questions and Solutions

Question 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 15 for the first km and Rs 8 for each additional km.

(ii). The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

(iii). 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.

(iv). The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:

To determine if a situation forms an Arithmetic Progression (AP), we need to check if the difference between consecutive terms is constant.

(i) Taxi Fare:

The fare for the first kilometer is Rs 15.

The fare for the first 2 kilometers is Rs 15 (for the 1st km) + Rs 8 (for the 2nd km) = Rs 23.

The fare for the first 3 kilometers is Rs 23 (for the first 2 km) + Rs 8 (for the 3rd km) = Rs 31.

The fare for the first 4 kilometers is Rs 31 (for the first 3 km) + Rs 8 (for the 4th km) = Rs 39.

The sequence of fares is 15, 23, 31, 39, ...

The difference between consecutive terms is: 23 - 15 = 8, 31 - 23 = 8, 39 - 31 = 8. Since the difference is constant (Rs 8), this situation forms an Arithmetic Progression.

(ii) Air in a Cylinder:

Let the initial volume of air in the cylinder be V.

After the first stroke, the pump removes ¼ of the air, so the remaining air is V - (1/4)V = (3/4)V.

After the second stroke, it removes ¼ of the remaining air: (3/4)V - (1/4)(3/4)V = (3/4)V * (1 - 1/4) = (3/4)V * (3/4) = (3/4)^2 V.

After the third stroke, the remaining air will be (3/4)^3 V.

The sequence of air volumes is V, (3/4)V, (3/4)^2 V, (3/4)^3 V, ...

The difference between consecutive terms is: (3/4)V - V = -(1/4)V, and (3/4)^2 V - (3/4)V = (9/16)V - (12/16)V = -(3/16)V. Since the differences are not constant, this situation does not form an Arithmetic Progression.

(iii) Cost of Digging a Well:

The cost for the first meter is Rs 150.

The cost for the first 2 meters is Rs 150 (for the 1st m) + Rs 50 (for the 2nd m) = Rs 200.

The cost for the first 3 meters is Rs 200 (for the first 2 m) + Rs 50 (for the 3rd m) = Rs 250.

The cost for the first 4 meters is Rs 250 (for the first 3 m) + Rs 50 (for the 4th m) = Rs 300.

The sequence of costs is 150, 200, 250, 300, ...

The difference between consecutive terms is: 200 - 150 = 50, 250 - 200 = 50, 300 - 250 = 50. Since the difference is constant (Rs 50), this situation forms an Arithmetic Progression.

(iv) Compound Interest:

The initial deposit is Rs 10000.

After 1 year, the amount will be $10000 \left(1 + \frac{8}{100}\right)$ .

After 2 years, the amount will be $10000 \left(1 + \frac{8}{100}\right)^2$ .

After 3 years, the amount will be $10000 \left(1 + \frac{8}{100}\right)^3$ .

The sequence of amounts is $10000, 10000\left(1+\frac{8}{100}\right), 10000\left(1+\frac{8}{100}\right)^2, 10000\left(1+\frac{8}{100}\right)^3, ...$

The ratio of consecutive terms is constant ( $1 + \frac{8}{100}$ ), indicating a Geometric Progression, not an Arithmetic Progression. The difference between consecutive terms is not constant. Therefore, this situation does not form an Arithmetic Progression.

Question 2

Write the first four terms of the A.P. when the first term a and the common difference d are given as follows:

$a = 10, d = 10$
$a = -2, d = 0$
$a = 4, d = -3$
$a = -1, d = \frac{1}{2}$
$a = -1.25, d = -0.25$

Solution:

The general form of an Arithmetic Progression (AP) is $a, a+d, a+2d, a+3d, ...$ , where 'a' is the first term and 'd' is the common difference. We need to find the first four terms for each given pair of 'a' and 'd'.

(i) $a = 10, d = 10$

First term ( $a_1$ ) = $a = 10$

Second term ( $a_2$ ) = $a + d = 10 + 10 = 20$

Third term ( $a_3$ ) = $a + 2d = 10 + 2(10) = 10 + 20 = 30$

Fourth term ( $a_4$ ) = $a + 3d = 10 + 3(10) = 10 + 30 = 40$

The first four terms are 10, 20, 30, 40.

(ii) $a = -2, d = 0$

First term ( $a_1$ ) = $a = -2$

Second term ( $a_2$ ) = $a + d = -2 + 0 = -2$

Third term ( $a_3$ ) = $a + 2d = -2 + 2(0) = -2 + 0 = -2$

Fourth term ( $a_4$ ) = $a + 3d = -2 + 3(0) = -2 + 0 = -2$

The first four terms are -2, -2, -2, -2.

(iii) $a = 4, d = -3$

First term ( $a_1$ ) = $a = 4$

Second term ( $a_2$ ) = $a + d = 4 + (-3) = 4 - 3 = 1$

Third term ( $a_3$ ) = $a + 2d = 4 + 2(-3) = 4 - 6 = -2$

Fourth term ( $a_4$ ) = $a + 3d = 4 + 3(-3) = 4 - 9 = -5$

The first four terms are 4, 1, -2, -5.

(iv) $a = -1, d = \frac{1}{2}$

First term ( $a_1$ ) = $a = -1$

Second term ( $a_2$ ) = $a + d = -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2}$

Third term ( $a_3$ ) = $a + 2d = -1 + 2\left(\frac{1}{2}\right) = -1 + 1 = 0$

Fourth term ( $a_4$ ) = $a + 3d = -1 + 3\left(\frac{1}{2}\right) = -1 + \frac{3}{2} = -\frac{2}{2} + \frac{3}{2} = \frac{1}{2}$

The first four terms are -1, -1/2, 0, 1/2.

(v) $a = -1.25, d = -0.25$

First term ( $a_1$ ) = $a = -1.25$

Second term ( $a_2$ ) = $a + d = -1.25 + (-0.25) = -1.25 - 0.25 = -1.50$

Third term ( $a_3$ ) = $a + 2d = -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75$

Fourth term ( $a_4$ ) = $a + 3d = -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00$

The first four terms are -1.25, -1.50, -1.75, -2.00.

Common mistakes

Confusing AP with other types of sequences (e.g., geometric progressions).
Incorrectly calculating the common difference.
Errors in applying the rule for subsequent terms in an AP.
Misinterpreting real-world scenarios to fit the AP definition.

Revision tips

Focus on understanding the condition for a sequence to be an AP: a constant common difference.
Practice identifying APs in diverse examples, both mathematical and practical.
Ensure you can correctly calculate the next term by adding the common difference.
Review the given 'a' and 'd' values carefully before calculating terms.

Practice MCQs

Q1. Which of the following situations describes an Arithmetic Progression?

Q2. If the first term 'a' is 10 and the common difference 'd' is 10, what is the third term of the AP?

Q3. What is the common difference in the cost of digging a well for each subsequent meter if the first meter costs Rs 150 and the second costs Rs 200?

Q4. A sequence is NOT an AP if:

Q5. If a = -2 and d = 0, what are the first four terms of the AP?

Frequently asked questions

What is an Arithmetic Progression (AP)?

An Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference (d).

How can I determine if a list of numbers forms an AP?

To check if a list of numbers forms an AP, calculate the difference between each pair of consecutive terms. If all these differences are the same, the list forms an AP.

What is the significance of the first term (a) and common difference (d) in an AP?

The first term (a) is the starting point of the sequence. The common difference (d) determines how each subsequent term is generated by adding 'd' to the previous term.

Are real-world examples provided in these solutions?

Yes, the solutions explain how concepts like taxi fares and digging costs can form an Arithmetic Progression, helping to understand the practical application of the topic.

How do these solutions help with exam preparation?

These solutions offer clear, step-by-step explanations for each problem, reinforcing understanding of AP concepts and methods, which is crucial for tackling exam questions effectively.

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