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NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 PDF

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions has been provided here. Students can refer to these solutions before their examination for better understanding.
authorImageNeha Tanna6 Jan, 2025
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NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1: Chapter 5 of Class 10 Maths, Arithmetic Progressions (APs), introduces students to sequences where each term increases or decreases by a fixed value. Exercise 5.1 identifies whether a given sequence forms an AP and determines its common difference, .

Students learn to analyze patterns and apply the AP formula for nth term, where a is the first term and is the term number. This foundational exercise helps develop logical reasoning and problem-solving skills, essential for tackling higher-order AP problems in the subsequent exercises.

Important Questions For Class 10 Maths Chapter 5

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Overview

Chapter 5 of Class 10 Maths, Arithmetic Progressions (APs) , is vital for understanding sequences with a constant difference between consecutive terms. Exercise 5.1 introduces the concept of APs and focuses on identifying sequences as APs and determining their common difference ( ).
Mastering this exercise builds a strong foundation for solving complex problems involving nth terms and sums of APs in later exercises. Understanding APs is not only crucial for academic success but also has practical applications in real-world scenarios, such as financial calculations, pattern recognition, and data analysis, making it an essential concept in both mathematics and daily life.

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 PDF

Chapter 5 of Class 10 Maths, Arithmetic Progressions, introduces the concept of sequences with a fixed difference between terms. Exercise 5.1 focuses on identifying APs and calculating their common difference, forming the foundation for advanced topics. To assist students, we have provided a comprehensive PDF of the NCERT Solutions for Exercise 5.1 below, offering clear explanations and step-by-step solutions to enhance understanding and support effective exam preparation.

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 PDF

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions

Below is the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions -

1. In which of the following situations does the list of numbers involved make an 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.

Solution:

We can write the given condition as Taxi fare for 1 km = 15 Taxi fare for first 2 kms = 15+8 = 23 Taxi fare for first 3 kms = 23+8 = 31 Taxi fare for first 4 kms = 31+8 = 39 And so on…… Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution:

Let the volume of air in a cylinder initially be V litres. In each stroke, the vacuum pump removes 1/4th of the remaining air in the cylinder at a time. Or we can say after every stroke, 1-1/4 = 3/4th part of the air will remain. Therefore, volumes will be V , 3 V /4 , (3 V /4) 2 , (3 V /4) 3 …and so on Clearly, we can see here the adjacent terms of this series do not have a common difference between them. Therefore, this series is not an A.P.

(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.

Solution:

We can write the given condition as; Cost of digging a well for first metre = Rs.150 Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200 Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250 Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300 And so on… Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.
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(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound interest at 8% per annum.

Solution:

We know that if Rs. P is deposited at r % compound interest per annum for n years, the amount of money will be P(1+r/100) n Therefore, after each year, the amount of money will be 10000(1+8/100), 10000(1+8/100) 2 , 10000(1+8/100) 3 …… Clearly, the terms of this series do not have a common difference between them. Therefore, this is not an A.P.

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

(i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = – 3 (iv) a = -1 d = 1/2 (v) a = – 1.25, d = – 0.25

Solutions:

(i) a = 10, d = 10 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = 10 a 2 = a 1 + d = 10+10 = 20 a 3 = a 2 + d = 20+10 = 30 a 4 = a 3 + d = 30+10 = 40 a 5 = a 4 + d = 40+10 = 50 And so on… Therefore, the A.P. series will be 10, 20, 30, 40, 50 … And the first four terms of this A.P. will be 10, 20, 30, and 40. (ii) a = – 2, d = 0 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = -2 a 2 = a 1 + d = – 2+0 = – 2 a 3 = a 2 +d = – 2+0 = – 2 a 4 = a 3 + d = – 2+0 = – 2 Therefore, the A.P. series will be – 2, – 2, – 2, – 2 … And the first four terms of this A.P. will be – 2, – 2, – 2 and – 2. (iii) a = 4, d = – 3 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = 4 a 2 = a 1 + d = 4-3 = 1 a 3 = a 2 + d = 1-3 = – 2 a 4 = a 3 + d = -2-3 = – 5 Therefore, the A.P. series will be 4, 1, – 2 – 5 … And the first four terms of this A.P. will be 4, 1, – 2 and – 5. (iv) a = – 1, d = 1/2 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 2 = a 1 + d = -1+1/2 = -1/2 a 3 = a 2 + d = -1/2+1/2 = 0 a 4 = a 3 + d = 0+1/2 = 1/2 Thus, the A.P. series will be-1, -1/2, 0, 1/2 And First four terms of this A.P. will be -1, -1/2, 0 and 1/2. (v) a = – 1.25, d = – 0.25 Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = – 1.25 a 2 = a 1 + d = – 1.25-0.25 = – 1.50 a 3 = a 2 + d = – 1.50-0.25 = – 1.75 a 4 = a 3 + d = – 1.75-0.25 = – 2.00 Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 …….. And the first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
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3. For the following A.P.s, write the first term and the common difference. (i) 3, 1, – 1, – 3 … (ii) -5, – 1, 3, 7 … (iii) 1/3, 5/3, 9/3, 13/3 …. (iv) 0.6, 1.7, 2.8, 3.9 …

Solutions

(i) Given series, 3, 1, – 1, – 3 … The first term, a = 3 The common difference, d = Second term – First term ⇒  1 – 3 = -2 ⇒  d = -2

(ii) Given series, – 5, – 1, 3, 7 …

The first term, a = -5 The common difference, d = Second term – First term ⇒ ( – 1)-( – 5) = – 1+5 = 4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….

The first term, a = 1/3 The common difference, d = Second term – First term ⇒ 5/3 – 1/3 = 4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 …

The first term, a = 0.6 The common difference, d = Second term – First term ⇒ 1.7 – 0.6 ⇒ 1.1

4. Which of the following are APs? If they form an A.P., find the common difference d and write three more terms.

(i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 …. (iii) -1.2, -3.2, -5.2, -7.2 … (iv) -10, – 6, – 2, 2 … (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 …. (vii) 0, – 4, – 8, – 12 … (viii) -1/2, -1/2, -1/2, -1/2 …. (ix) 1, 3, 9, 27 … (x) a , 2 a , 3 a , 4 a … (xi) a , a 2 , a 3 , a 4 … (xii) √2, √8, √18, √32 … (xiii) √3, √6, √9, √12 … (xiv) 1 2 , 3 2 , 5 2 , 7 2 … (xv) 1 2 , 5 2 , 7 2 , 7 3 …

Solution

(i) Given to us, 2, 4, 8, 16 … Here, the common difference is a 2 – a 1 = 4 – 2 = 2 a 3 – a 2 = 8 – 4 = 4 a 4 – a 3 = 16 – 8 = 8 a n +1 – a n or the common difference is not the same every time. Therefore, the given series is not forming an A.P.

(ii) Given, 2, 5/2, 3, 7/2 ….

Here, a 2 – a 1 = 5/2-2 = 1/2 a 3 – a 2 = 3-5/2 = 1/2 a 4 – a 3 = 7/2-3 = 1/2 a n +1 – a n or the common difference is the same every time. Therefore, d = 1/2 and the given series are in A.P. The next three terms are a 5 = 7/2+1/2 = 4 a 6 = 4 +1/2 = 9/2 a 7 = 9/2 +1/2 = 5

(iii) Given, -1.2, – 3.2, -5.2, -7.2 …

Here, a 2 – a 1 = (-3.2)-(-1.2) = -2 a 3 – a 2 = (-5.2)-(-3.2) = -2 a 4 – a 3 = (-7.2)-(-5.2) = -2 a n +1 – a n or common difference is the same every time. Therefore, d = -2 and the given series are in A.P. Hence, the next three terms are a 5 = – 7.2-2 = -9.2 a 6 = – 9.2-2 = – 11.2 a 7 = – 11.2-2 = – 13.2

(iv) Given, -10, – 6, – 2, 2 …

Here, the terms and their difference are a 2 – a 1 = (-6)-(-10) = 4 a 3 – a 2 = (-2)-(-6) = 4 a 4 – a 3 = (2 -(-2) = 4 a n +1 – a n or the common difference is the same every time. Therefore, d = 4, and the given numbers are in A.P. Hence, the next three terms are a 5 = 2+4 = 6 a 6 = 6+4 = 10 a 7 = 10+4 = 14

(v) Given, 3, 3+√2, 3+2√2, 3+3√2

Here, a 2 – a 1 = 3+√2-3 = √2 a 3 – a 2 = (3+2√2)-(3+√2) = √2 a 4 – a 3 = (3+3√2) – (3+2√2) = √2 a n +1 – a n or the common difference is the same every time. Therefore, d = √2, and the given series forms an A.P. Hence, the next three terms are a 5 = (3+√2) +√2 = 3+4√2 a 6 = (3+4√2)+√2 = 3+5√2 a 7 = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Here, a 2 – a 1 = 0.22-0.2 = 0.02 a 3 – a 2 = 0.222-0.22 = 0.002 a 4 – a 3 = 0.2222-0.222 = 0.0002 a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.

(vii) 0, -4, -8, -12 …

Here, a 2 – a 1 = (-4)-0 = -4 a 3 – a 2 = (-8)-(-4) = -4 a 4 – a 3 = (-12)-(-8) = -4 a n +1 – a n or the common difference is the same every time. Therefore, d = -4 and the given series forms an A.P. Hence, the next three terms are a 5 = -12-4 = -16 a 6 = -16-4 = -20 a 7 = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Here, a 2 – a 1 = (-1/2) – (-1/2) = 0 a 3 – a 2 = (-1/2) – (-1/2) = 0 a 4 – a 3 = (-1/2) – (-1/2) = 0 a n +1 – a n or the common difference is the same every time. Therefore, d = 0 and the given series forms an A.P. Hence, the next three terms are; a 5 = (-1/2)-0 = -1/2 a 6 = (-1/2)-0 = -1/2 a 7 = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

Here, a 2 – a 1 = 3-1 = 2 a 3 – a 2 = 9-3 = 6 a 4 – a 3 = 27-9 = 18 a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.

(x) a , 2 a , 3 a , 4 a …

Here, a 2 – a 1 = 2 a – a = a a 3 – a 2 = 3 a -2 a = a a 4 – a 3 = 4 a -3 a = a a n +1 – a n or the common difference is the same every time. Therefore, d = a and the given series forms an A.P. Hence, the next three terms are a 5 = 4 a + a = 5 a a 6 = 5 a + a = 6 a a 7 = 6 a + a = 7 a

(xi) a , a 2 , a 3 , a 4 …

Here, a 2 – a 1 = a 2 – a = a( a -1) a 3 – a 2 = a 3 – a 2 = a 2 ( a -1) a 4 – a 3 = a 4 – a 3 = a 3 ( a -1) a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.

(xii) √2, √8, √18, √32 …

Here, a 2 – a 1 = √8-√2  = 2√2-√2 = √2 a 3 – a 2 = √18-√8 = 3√2-2√2 = √2 a 4 – a 3 = 4√2-3√2 = √2 a n +1 – a n or the common difference is the same every time. Therefore, d = √2, and the given series forms an A.P. Hence, the next three terms are a 5 = √32+√2 = 4√2+√2 = 5√2 = √50 a 6 = 5√2+√2 = 6√2 = √72 a 7 = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Here, a 2 – a 1 = √6-√3 = √3×√2-√3 = √3(√2-1) a 3 – a 2 = √9-√6 = 3-√6 = √3(√3-√2) a 4 – a 3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3) a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.

(xiv) 1 2 , 3 2 , 5 2 , 7 2 …

Or, 1, 9, 25, 49 ….. Here, a 2 − a 1 = 9−1 = 8 a 3 − a 2 = 25−9 = 16 a 4 − a 3 = 49−25 = 24 a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.

(xv) 1 2 , 5 2 , 7 2 , 73 …

Or 1, 25, 49, 73 … Here, a 2 − a 1 = 25−1 = 24 a 3 − a 2 = 49−25 = 24 a 4 − a 3 = 73−49 = 24 a n +1 – a n or the common difference is the same every time. Therefore, d = 24 and the given series forms an A.P. Hence, the next three terms are a 5 = 73+24 = 97 a 6 = 97+24 = 121 a 7 = 121+24 = 145

Benefits of Using NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Clear Understanding: Solutions offer step-by-step explanations, helping students grasp the concept of Arithmetic Progressions and their application effectively.

Exam Preparation: Practicing with NCERT solutions ensures alignment with the exam syllabus and pattern.

Problem-Solving Skills: Builds confidence by providing detailed methods to solve AP-related questions.

Time Management: Familiarity with solution techniques improves efficiency during exams.

Concept Clarity: Reinforces foundational concepts, essential for advanced mathematical studies and real-life applications.

Self-Assessment: Allows students to check and improve their problem-solving accuracy.

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 FAQs

What is the purpose of arithmetic progression?

The purpose of arithmetic progression is to identify patterns in sequences with a constant difference, enabling problem-solving in mathematics, data analysis, and real-world applications like finance and scheduling.

What is the average of the arithmetic progression?

Let us say that we have an A. P. which has several evenly spaced numbers. Let the first term of the sequence be x1 and the last term be xn. Then the average of the series can be found out by the following formula: Average = (x1 + xn)/2.

How to check if arithmetic progression?

The first term of an arithmetic progression is usually denoted by 'a' or 'a1'.

Who discovered arithmetic progression?

Johann Carl Friedrich Gauss is the father of Arithmetic Progression. He found it when he was in school and his teacher asked to sum the integers from 1 to 100.
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