# NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1

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## NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progression Ex 5.1

CBSE students of Class 10 can easily download solutions of Class 10 Maths Chapter 5 Exercise 5.1. They can download the NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 from Extramarks’ website or mobile application in PDF format so that they can study the NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 even in offline mode.

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### Access NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progression

All the tools required for competitive exam preparation are available on the Extramarks website and mobile application. Students can use Extramarks to help them get ready for a variety of competitive exams, including JEE Mains, NEET, JEE Advance, CUET, and others. Extramarks provides all the solutions to every chapter and subject. Students can go to the website of Extramarks and access the NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 in PDF format for further assistance regarding that chapter.

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### NCERT Solutions for Class 10 Maths – Exercise 5.1

There are a total of 15 chapters in Class 10, and each chapter is important for the understanding of students. Students must go through each chapter and answer every question in the chapters. The NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 will help out students whenever they get stuck on any question from the exercise. Extramarks provides solutions for every chapter and exercise of Class 10 and more. Students must have a clear concept while solving any question. They are advised to go through the syllabus of Class 10 Mathematics and then organise their study schedule accordingly. There are a total of 4 questions in exercise 5.1 of Chapter Arithmetic Progression. As question number one has four questions, each of which has four sub-questions.Then question 2 has 5 questions, question 3 has 4 questions, and finally question 4 has 15 questions. The exercise may seem short, but it has many sub-questions that could be challenging for students to solve, and NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 will help students solve every question. Students must attempt to solve each question independently after consulting NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1.The methods and solutions are the easiest available on the internet. Students can follow the steps of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 to score better in examinations. Students will see a rise in their marks if they follow the NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1. Students are advised to take each and every exercise seriously and solve them again and again until they have a clear understanding of them. Students will definitely score well if they follow the routine and give proper time to every subject and topic.

### NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1

Students can download the free PDF of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1. They will get the PDF of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 from Extramarks’ website or mobile application so that they can study the materials even in offline mode. Students must not get distracted while solving the questions in the NCERT Mathematics book and solve those questions with full attention.The NCERT Solutions for Class 10 Maths, Chapter 5, Exercise 5.1, are available for free on the Extramarks website or through the Extramarks app.

Downloading the PDF format of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 will be helpful for students while solving the questions. Students at times get baffled while solving questions; hence, they must remember that NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 will solve all their queries. Students can download the PDF format of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 by clicking on the link below.

An Overview of the Topics Covered under NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions (Exercise 5.1)

The beginning of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions provides a few sequence examples to aid students in recognising the various number patterns. The chapter goes into more detail on the terms, common differences, and other terms used in the arithmetic progression. The students will learn that an arithmetic progression is a sequence of numbers where each term can be found by adding a fixed value to the preceding term, leaving aside the first term. The “common difference” is a set value or number that might be positive, negative, or zero. The students must keep in mind that an arithmetic progression often takes the following form: a, a + d, a + 2d, a + 3d,…, where’d’ is a common difference. This NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 demonstrates the differences between finite and infinite arithmetic progression using pertinent examples. A finite AP has the last term, whereas an infinite AP does not.

There are 4 questions in this exercise that thoroughly cover the topic. The students are expected to identify frequent differences and AP sequences, as well as the first few AP terms for the supplied first term. The following link will take students to the pdf of the NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1.

### Importance of Class 10 Maths Chapter 5 Arithmetic Progression

The difference between any two consecutive numbers in an arithmetic progression is always the constant value. A typical difference is a name given to the constant value.

The practical use of arithmetic progression:

To figure out how many spectators a stadium can accommodate.

When we take a taxi, we will be charged an initial cost and then a per mile or kilometre charge.

This chapter will help students score well in their examinations, and understanding this chapter will help them use it in their daily lives as well. Therefore, students must solve each and every question in this chapter to prevent being clueless in the examination. The NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 could be a treasure for students in their time of need. That is why students must practise the problems of Arithmetic Progression.

### NCERT Solution Class 10 Maths Chapter 5 – Arithmetic Progressions Exercises

As Arithmetic Progression is one of the most important chapters of Class 10. So, students must solve every exercise of the given chapter and Extramarks has provided every solution to each question of Chapter 5- Arithmetic Progression. There are a total of 4 exercises in the given chapter ,and each exercise has been solved by the experts of Mathematics. The first exercise 5.1 has a total number of 4 questions and the second exercise 5.2 has a total number of 20 questions, the third exercise 5.3 has a total number of 20 questions as well, and lastly, the 5.4 exercise is the optional one but students are advised to solve it too. Exercise 5.4 has a total number of 5 questions. Students are advised to solve each and every question in the chapter so that they will not face any issues while solving questions in the examination hall. Extramarks has solved each exercise; the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 are available on the website.Then NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.2, NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3, and NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.4, are available on the website of Extramarks and mobile application as well. Students may follow the steps of NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 to have a better understanding of the exercise and chapter.

### Class 10 Maths Weightage Marks

The Central Board of Secondary Education’s (CBSE) Class 10 Mathematics curriculum is now accessible. Students can get the Class 10 Mathematics Syllabus for the academic year 2022–2023 from the Extramarks website. The CBSE Mathematics Class 10 Syllabus 2022–23 lists the unit names, themes, and marks assigned for each unit.

Students should complete the class 10 Mathematics curriculum for the years 2022–2032 at least a month before the CBSE Board 10th examinations in 2023. Students should finish the CBSE Class 10 Mathematics Syllabus 2023 as soon as they can. The theory paper of Mathematics will be 80 marks each. Read this article to download the CBSE class 10 Mathematics syllabus 2023 and know the exam pattern and preparation tips.

### Benefits of Ex 5.1 Class 10 Maths Solutions

The concept of arithmetic progressions is covered in Chapter 5 of the Class 10 Mathematics NCERT text, which is a particularly fascinating subject. The questions in this chapter can all be quite demanding. Hence, students would have to think logically while solving each question, which is why this might very well be one of the favourite chapters for most students. Students must properly comprehend the chapter with the aid of the Extramarks NCERT solutions. This way, students will be able to have a clear understanding and enjoy their study time simultaneously. Students must know that Exercise 1.5 is one of the important exercises of the chapter. The NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.1 will help students solve the questions immediately.

Students can also download NCERT Solutions for other classes by clicking the links given below-

NCERT Solutions Class 11

NCERT Solutions class 10

NCERT Solutions Class 9

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NCERT Solutions Class 7

NCERT Solutions class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

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NCERT Solutions Class 2

NCERT Solutions Class 1

**Q.1 **

$\begin{array}{l}\text{Write first four terms of the AP, when the first term}\\ \text{a and the common difference d are given as follows:}\\ \text{(i)}\mathrm{a}=10,\text{}\mathrm{d}=10\text{}\\ \text{(ii)}\mathrm{a}=-2,\text{}\mathrm{d}=0\\ \text{(iii)}\mathrm{a}=4,\text{}\mathrm{d}=-3\text{}\\ \text{(iv)}\mathrm{a}=-1,\text{}\mathrm{d}=\frac{1}{2}\\ \text{(v)}\mathrm{a}=-1.25,\text{}\mathrm{d}=-0.25\end{array}$

**Ans.**

$\begin{array}{l}\text{(i)}\\ \text{We have,}\\ \text{First term}=\text{a}=10\text{}\\ \text{and}\\ \text{Common difference}=\text{d}=10\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=\text{10}+10=20,\\ \text{Third term}=\text{a}+2\text{d}=\text{10}+2\times 10=30,\\ \text{Fourth term}=\text{a}+3\text{d}=\text{10}+3\times 10=40,\\ \text{Therefore, first four terms of the AP are:}\\ \text{10, 20, 30, 40}\\ \text{(ii)}\\ \text{We have,}\\ \text{First term}=\text{a}=-2\\ \text{and}\\ \text{Common difference}=\text{d}=0\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-2+0=-2,\\ \text{Third term}=\text{a}+2\text{d}=-2+2\times 0=-2,\\ \text{Fourth term}=\text{a}+3\text{d}=-2+3\times 0=-2,\\ \text{Therefore, first four terms of the AP are:}\\ -2\text{,}-2\text{,}-2\text{,}-2\\ \text{(iii)}\\ \text{We have,}\\ \text{First term}=\text{a}=4\\ \text{and}\\ \text{Common difference}=\text{d}=-3\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=4-3=1,\\ \text{Third term}=\text{a}+2\text{d}=4+2\times (-3)=-2,\\ \text{Fourth term}=\text{a}+3\text{d}=4+3\times (-3)=-5,\\ \text{Therefore, first four terms of the AP are:}\\ 4\text{,}1\text{,\hspace{0.17em}\hspace{0.17em}}-2\text{,\hspace{0.17em}\hspace{0.17em}}-5\\ \\ \text{(iv)}\\ \text{We have,}\\ \text{First term}=\text{a}=-1\\ \text{and}\\ \text{Common difference}=\text{d}=\frac{1}{2}\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-1+\frac{1}{2}=-\frac{1}{2}\\ \text{Third term}=\text{a}+2\text{d}=-1+2\times \frac{1}{2}=0\\ \text{Fourth term}=\text{a}+3\text{d}=-1+3\times \frac{1}{2}=\frac{1}{2}\\ \text{Therefore, first four terms of the AP are:}\\ -\text{1,}-\frac{1}{2},\text{0,}\frac{1}{2}\\ \text{(v)}\\ \text{We have,}\\ \text{First term}=\text{a}=-1.25\\ \text{and}\\ \text{Common difference}=\text{d}=-0.25\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-1.25-0.25=-1.50,\\ \text{Third term}=\text{a}+2\text{d}=-1.25+2\times (-0.25)=-1.75,\\ \text{Fourth term}=\text{a}+3\text{d}=-1.25+3\times (-0.25)=-2,\\ \text{Therefore, first four terms of the AP are:}\\ -1.25\text{,}-1.50\text{,\hspace{0.17em}\hspace{0.17em}}-1.75\text{,\hspace{0.17em}\hspace{0.17em}}-2\end{array}$

**Q.2 **

$\begin{array}{l}\text{For the following APs, write the first term and the}\\ \text{common difference:}\\ \text{(i)}3,\text{\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}-3,\text{}.\text{}.\text{}.\text{(ii)}-\text{5,\hspace{0.17em}\hspace{0.17em}}-\text{1, 3, 7, . . .}\\ \text{(iii)}\frac{\text{1}}{3}\text{,\hspace{0.17em}\hspace{0.17em}}\frac{\text{5}}{3}\text{,}\frac{\text{9}}{3}\text{,\hspace{0.17em}\hspace{0.17em}}\frac{\text{13}}{3}\text{,. . . (iv) 0.6, 1.7, 2.8, 3.9, . . .}\end{array}$

**Ans.**

$\begin{array}{l}\text{(i)}\\ \text{Following is the given AP.}\\ \text{}3,\text{\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}-3,\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=3\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-3=-2\\ \text{(ii)}\\ \text{Following is the given AP.}\\ \text{}-5,\text{\hspace{0.17em}\hspace{0.17em}}-1,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}7,\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=-5\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1-5=-6\\ \\ \text{(iii)}\\ \text{Following is the given AP.}\\ \text{}\frac{1}{3},\text{\hspace{0.17em}\hspace{0.17em}}\frac{5}{3},\text{}\frac{9}{3},\text{}\frac{13}{9},\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=\frac{1}{3}\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\\ \\ \text{(iv)}\\ \text{Following is the given AP.}\\ \text{0.6, 1.7, 2.8, 3.9,}.\text{}.\text{}.\text{}\\ \text{Here, first term}=0.6\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.7-0.6=\text{1.1}\end{array}$

**Q.3 **

$\begin{array}{l}\text{Which of the following are APs? If they form an AP,}\\ \text{find the common difference d and write three more}\\ \text{terms.}\\ \text{(i) 2, 4, 8, 16, . . .}\\ \text{(ii) 2,}\frac{\text{5}}{2}\text{, 3,}\frac{\text{7}}{2}\text{, . . .}\\ \text{(iii)}-\text{1.2,}-\text{3.2,}-\text{5.2,}-\text{7.2, . . .}\\ \text{(iv)}-\text{10,}-\text{6,}-\text{2, 2, . . .}\\ \text{(v) 3,}3+\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+2\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+3\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{(vi)}0.2,\text{}0.22,\text{}0.222,\text{}0.2222,\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{(vii) 0,}-\text{4,}-\text{8,}-\text{12, . . .}\\ \text{(viii)}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{, . . .}\\ \text{(ix) 1, 3, 9, 27,}...\\ \text{(x) a, 2a, 3a, 4a,}...\\ {\text{(xi) a, a}}^{2},{\text{a}}^{3},{\text{a}}^{4},\text{}...\\ \text{(xii)}\sqrt{2},\text{}\sqrt{8},\text{}\sqrt{18},\text{}\sqrt{32},\text{}...\\ \text{(xiii)}\sqrt{3},\text{}\sqrt{6},\text{}\sqrt{9},\text{}\sqrt{12},\text{}...\\ {\text{(xiv) 1}}^{2}{\text{, 3}}^{2},{\text{5}}^{2},{\text{7}}^{2},\text{}...\\ {\text{(xv) 1}}^{2}{\text{, 5}}^{2},{\text{7}}^{2},\text{73},\text{}...\end{array}$

**Ans.**

$\begin{array}{l}\text{(i) 2, 4, 8, 16, . . .}\\ \text{Here, 4}-2=2\ne 8-4\text{}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(ii) 2,}\frac{\text{5}}{2}\text{, 3,}\frac{\text{7}}{2}\text{, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\frac{\text{5}}{2}-2=\frac{1}{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=\text{3}-\frac{\text{5}}{2}=\frac{1}{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=\frac{\text{7}}{2}-3=\frac{1}{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\frac{1}{2}.\\ \text{The next three terms are:}\frac{\text{7}}{2}+\frac{1}{2}=4,\text{\hspace{0.17em}\hspace{0.17em}}4+\frac{1}{2}=\frac{9}{2}\text{and}\frac{9}{2}+\frac{1}{2}=5.\\ \text{(iii)}-\text{1.2,}-\text{3.2,}-\text{5.2,}-\text{7.2, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-\text{3.2}-(-\text{1.2})=-\text{3.2}+\text{1.2}=-2\\ {\text{a}}_{3}-{\text{a}}_{2}=-\text{5.2}-(-3.\text{2})=-\text{5.2}+3.\text{2}=-2\\ {\text{a}}_{4}-{\text{a}}_{3}=-7.\text{2}-(-5.\text{2})=-7.\text{2}+5.\text{2}=-2\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=-2.\\ \text{The next three terms are:}-\text{7.2}-2=-9.\text{2},\text{\hspace{0.17em}\hspace{0.17em}}-9.\text{2}-2=-11.\text{2}\\ \text{and}-11.\text{2}-2=-13.\text{2}.\\ \text{(iv)}-\text{10,}-\text{6,}-\text{2, 2, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-\text{6}-(-\text{10})=-\text{6}+\text{10}=4\\ {\text{a}}_{3}-{\text{a}}_{2}=-\text{2}-(-6)=-\text{2}+6=4\\ {\text{a}}_{4}-{\text{a}}_{3}=\text{2}-(-\text{2})=\text{2}+\text{2}=4\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=4.\\ \text{The next three terms are: 2}+4=6,\text{\hspace{0.17em}\hspace{0.17em}}6+4=10\text{and}10+4=14.\\ \text{(v) 3,}3+\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+2\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+3\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=3+\sqrt{2}-3=\sqrt{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=3+2\sqrt{2}-(3+\sqrt{2})=\sqrt{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=3+3\sqrt{2}-(3+2\sqrt{2})=\sqrt{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\sqrt{2}.\\ \text{The next three terms are:}3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\text{and}3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}.\\ \text{(vi)}0.2,\text{}0.22,\text{}0.222,\text{}0.2222,\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=0.22-0.2=0.02\\ {\text{a}}_{3}-{\text{a}}_{2}=0.222-0.22=0.002\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(vii) 0,}-\text{4,}-\text{8,}-\text{12, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-4-\text{0}=-4\\ {\text{a}}_{3}-{\text{a}}_{2}=-8-(-4)=-8+4=-4\\ {\text{a}}_{4}-{\text{a}}_{3}=-1\text{2}-(-8)=-1\text{2}+\text{8}=-4\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=-4.\\ \text{The next three terms are:}-1\text{2}-4=-16,\text{\hspace{0.17em}\hspace{0.17em}}-16-4=-20\\ \text{and}-20-4=-24.\\ \text{(viii)}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,\hspace{0.17em}\hspace{0.17em}. . .}\\ {\text{Obviously, a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=0.\\ \text{The next three terms are:}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}.\\ \text{(ix) 1, 3, 9, 27,}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=3-1=2\\ {\text{a}}_{3}-{\text{a}}_{2}=9-3=6\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(x) a, 2a, 3a, 4a,}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=2\text{a}-\text{a}=\text{a}\\ {\text{a}}_{3}-{\text{a}}_{2}=\text{3a}-2\text{a}=\text{a}\\ {\text{a}}_{4}-{\text{a}}_{3}=\text{4a}-\text{3a}=\text{a}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\text{a}.\\ \text{The next three terms are:}4\text{a}+\text{a}=5\text{a},\text{\hspace{0.17em}\hspace{0.17em}}5\text{a}+\text{a}=6\text{a}\\ \text{and 6a}+\text{a}=7\text{a}.\\ {\text{(xi) a, a}}^{2},{\text{a}}^{3},{\text{a}}^{4},\text{}...\\ {\text{Obviously, a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is not the same every time.}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(xii)}\sqrt{2},\text{}\sqrt{8},\text{}\sqrt{18},\text{}\sqrt{32},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=\sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=\sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\sqrt{2}.\\ \text{The next three terms are:}\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\sqrt{50}+\sqrt{2}=5\sqrt{2}+\sqrt{2}=6\sqrt{2}=\sqrt{72}\text{}\\ \text{and}\sqrt{72}+\sqrt{2}=6\sqrt{2}+\sqrt{2}=7\sqrt{2}=\sqrt{98}.\\ \text{(xiii)}\sqrt{3},\text{}\sqrt{6},\text{}\sqrt{9},\text{}\sqrt{12},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\sqrt{6}-\sqrt{3}=\sqrt{2}\times \sqrt{3}-\sqrt{3}=\sqrt{3}\left(\sqrt{2}-1\right)\\ {\text{a}}_{3}-{\text{a}}_{2}=\sqrt{9}-\sqrt{6}=\sqrt{3}\times \sqrt{3}-\sqrt{2}\times \sqrt{3}=\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ {\text{(xiv) 1}}^{2}{\text{, 3}}^{2},{\text{5}}^{2},{\text{7}}^{2},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}={\text{3}}^{2}-{\text{1}}^{2}=9-1=8\\ {\text{a}}_{3}-{\text{a}}_{2}={\text{5}}^{2}-{\text{3}}^{2}=25-9=16\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ {\text{(xv) 1}}^{2}{\text{, 5}}^{2},{\text{7}}^{2},\text{73},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}={5}^{2}-{\text{1}}^{2}=25-1=24\\ {\text{a}}_{3}-{\text{a}}_{2}={\text{7}}^{2}-{\text{5}}^{2}=49-25=24\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{4}-{\text{a}}_{3}=\text{73}-{\text{7}}^{2}=73-49=24\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=24.\\ \text{The next three terms are: 73}+24=97,\text{\hspace{0.17em}}97+24=121\\ \text{and}121+24=145.\end{array}$

**Q.4 ** Fill in the blanks in the following table, given that *a* is the first term, *d* the common difference and *a _{n}* the nth term of the AP:

**Ans.**

$\begin{array}{l}\text{(i)}\\ \text{We have}\\ \mathrm{a}=7,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=8\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow {\mathrm{a}}_{\mathrm{n}}=7+(8-1)3=28\\ \text{(ii)}\\ \text{We have}\\ \mathrm{a}=-18,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=10,\text{}{\mathrm{a}}_{\mathrm{n}}=0\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow 0=-18+(10-1)\mathrm{d}\\ \Rightarrow \mathrm{d}=\frac{18}{9}=2\\ \text{(iii)}\\ \text{We have}\\ \mathrm{d}=-3,\text{}\mathrm{n}=18,\text{}{\mathrm{a}}_{\mathrm{n}}=-5\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow -5=\mathrm{a}+(18-1)(-3)\\ \Rightarrow \mathrm{a}=51-5=46\\ \text{(iv)}\\ \text{We have}\\ \mathrm{a}=-18.9,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=2.5,\text{}{\mathrm{a}}_{\mathrm{n}}=3.6\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow 3.6=-18.9+(\mathrm{n}-1)(2.5)\\ \Rightarrow (\mathrm{n}-1)(2.5)=3.6+18.9=22.5\\ \Rightarrow \mathrm{n}=\frac{22.5}{2.5}+1=10\\ \text{(v)}\\ \text{We have}\\ \mathrm{a}=3.5,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=105\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}\\ \Rightarrow {\mathrm{a}}_{\mathrm{n}}=3.5+(105-1)\times 0=3.5\end{array}$

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