NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

Home » NCERT Solutions » NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

CBSE

›

›

›

›

›

›

NEET 2023 - Exam Dates, Syllabus, Eligibility and Preparation Tips

›

›

›

›

›

NCERT books are the main textbooks for students of the CBSE board, and with their optimum use, students can be adequately prepared for their future endeavours. These books are incredibly beneficial to students and provide numerous advantages related to the strengthening of their basic knowledge. NCERT books help to develop a comprehensive understanding of the fundamental aspects of each subject in a comprehensive manner. Students can understand the concepts adequately by thoroughly reading NCERT books. This is bound to be beneficial in the long run since students would be able to solve problems effectively and efficiently once they have the necessary conceptual clarity. If students are able to understand concepts, then they will not have to retain them. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help out students when they have any queries. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are specially designed for students who are planning to appear in competitive exams. Students preparing for board exams in Class 12 begin their preparation in Class 11, and the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will play an important role in this regard.

Mathematics may seem like a complex academic discipline, but students ought to be aware of the fact that it includes the contributions of great mathematicians from various parts of the world, compiled over centuries. NCERT books are designed with students’ ages in mind, taking into account what their minds can comprehend at that age.The intellectual development of students is the objective of the NCERT textbooks. Keeping in mind the pattern of NCERT books, the Extramarks online learning platform has compiled NCERT Solutions to aid students in their academic pursuits. The use of Mathematics has spread to other subjects too, such as Physics, Chemistry, Economics, Engineering etc. These subjects need Mathematics to understand certain topics. The relevance of Mathematics is present everywhere, from an art museum to the home. The application of Mathematics is implicit in forms, shapes, sizes, degrees, temperatures, finance, time, and watches. Mathematics is offered as an ‘instrumental subject’ in some curricula to supplement the study of other school subjects, while in others, integrated courses that combine mathematics and other fields are offered.

The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are key materials for students who are planning to appear in different competitive examinations. NCERT books help students in developing a clear-cut understanding of various topics to ensure that they will be able to solve all types of questions in the board examinations as well as the annual examinations. Therefore, Extramarks provides solutions to a number of NCERT books so that students will not face problems while solving questions. NCERT is the apex body for the management of the school education system and academic curriculum in the country. Therefore, students belonging to the CBSE board must be aware of the fact that scoring well in CBSE examinations could be rendered possible by adequately preparing the books of NCERT. Extramarks, as an online learning platform for students, provides solutions to all those questions that are in NCERT books and more. There is no doubt that the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been provided for students to score well and learn smart and simple steps for solving the questions of Mathematics.

Students who study minutely from NCERT books will ace the examinations and would have their doubts clarified for examinations. So, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are bound to be useful for the times when students have recurring doubts while attempting to solve a particular type of question. To enable students to overcome all these hindrances, Extramarks could prove to be indispensable for students and could improve their conceptual understanding for the better. By solving problems with the aid of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, students would be able to achieve efficiency. They would be able to solve similar problems quite easily, as the format of these solutions has been prepared in a way that makes them easy to understand and time-saving. Extramarks has left no stone unturned to make students understand each topic of every subject easily.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

Students of Class 11 can easily download the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 from the Extramarks website or its mobile application. With the aid of this excellent resource for learning, students can gain access to solutions of Exercise 9.3 Class 11th in PDF format. They can refer to and revise with the aid of these resources even when they are not connected to the internet. The solutions to Class 11th Exercise 9.3 will help them understand questions better.

The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would be helpful for Class 11 Chapter 9- Sequences and Series. Class 11 occupies an important position in the academic career of a student and is an important stepping stone toward higher studies. If students follow the guidance of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 provided by Extramarks, then they will not be hindered by unresolved queries and questions. Extramarks abides by the pattern of the prescribed NCERT textbook for Mathematics. Class 11 is a decisive year for the academic pursuits of students. Students must have clarity in terms of important concepts, and they should be aware of the benefits and challenges of the streams of their choice. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would address the challenges and issues faced by students to a large extent. Class 10 gives students a broad understanding of every subject and after that, they have to decide which subjects pique their interest the most. Mathematics serves many purposes in day-to-day life. So, it is important to be conscious of the application of Mathematics. Therefore, opting for Mathematics in Class 11 would be a favourable decision for students. Students must be careful while selecting their subject stream since their future academic pursuits largely depend on this choice. Students must have a clear idea of what field of study they are willing to strive for in the future. The mentoring services provided by Extramarks could prove to be of great assistance in this regard.

Those students who choose the Science stream for their further education must know that this choice will help them grow. Students should solve Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, on a regular basis in order to understand the exercise clearly. A large proportion of the questions that appear in the examinations are derived from the NCERT textbooks. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help students with both understanding the questions as well as with regard to the application of concepts for solving said questions. Students ought to know how to apply certain concepts to seemingly complex questions, and they will be able to do so only when they practice solving similar questions on a routine basis. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 contain very easy solutions to simplify conceptual understanding for students. Therefore, students should seek the help of Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 to gain a flawless understanding of significant concepts.

Access NCERT Solutions for Class 11 Maths Chapter 9 –Sequences and Series

Students can gain access to the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 from the Extramarks educational website and mobile application. This application is very convenient to use, even for those students who are planning to appear in competitive examinations like the Joint Entrance Exams (JEE). These solutions will help in laying a reliable academic foundation for students who are determined to secure admission to prestigious institutions like the Indian Institute of Technology (IITs) and the National Institute of Technology (NITs). Additionally, these learning assets could also aid students who are aiming to pass examinations like the Common Universities Entrance Test (CUET) with merit. All of these examinations have a distinct pattern of assessment and a challenging schema of marks distribution. The questions are in the form of multiple choice questions where one or sometimes even more than one option could be correct. Moreover, there is often a provision of negative marking for wrong answers. Such an examination pattern necessitates decisive choices and single-minded concentration, which could be achieved through conceptual clarity and perseverance. Hard work and practise are the two keystones of a strong academic foundation, and students are greatly recommended to be sincere in their efforts. The online learning resources provided by the Extramarks educational website would complement the efforts of students towards the fulfilment of their academic objectives.

The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will be of great help to the students. Mathematics is an important academic discipline to qualify for competitive exams like the Joint Entrance Examination (JEE). The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been fashioned in a framework that would help students enhance their memory retention capabilities with regard to important formulae and concepts. Extramarks is a learning platform where students can easily access the solutions of Class 11 in PDF format, so it would be helpful even in offline mode. Accessing the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would be favourable for improving students’ understanding. Class 11 proves to be challenging for a majority of students because the syllabus gets incredibly vast after Class 10. Students find it difficult to cope with a sudden increase in the academic curriculum.

Students must know that Extramarks is a platform where students are instructed and guided by professional mentors in all subjects. They must gain access to the Extramarks educational portal so that they can easily refer to the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. Accordingly, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are easily available for students, and they can effectively rely on Extramarks for the resolution of their doubts and queries.

Students should comply with the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 since these solutions have been prepared by Extramarks with the collaboration of experienced experts in Mathematics. Once students have acquired a certain extent of understanding and clarity with regard to the concepts and themes of the prescribed curriculum of Mathematics for Class 11, they would be able to perform well in the examinations. Additionally, they will also be able to figure out their own strongholds in the subject, which would then be crucial while they are deciding on a career field to pursue as a part of their future academic or professional ventures. Students are advised to study regularly and revise diligently. TThey are also advised to seek advice and support from their respective teachers and mentors in order to resolve any recurring issues.They must maintain contact with the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 because Extramarks will always be available to help them.The Science stream could prove to be an insurmountable challenge at times, but students must seek the help of Extramarks. The academic curriculum of Class 11 provides in-depth knowledge of specific streams that students may choose.Students ought to remember that they will be appearing for the board examinations of Class 12 in the imminent future, and so they will need to be adequately prepared. Therefore, students must understand that studying diligently is the best solution for achieving excellence in the board examinations. Regular practise and self-study would serve to boost confidence, and revision would help students determine their next step forward decisively. The academic curriculum presented to students consists of many interesting and intriguing themes and concepts, which would undoubtedly serve as food for thought for the students. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help students at every step as they make their way towards achieving excellence. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would make the academic discipline of Mathematics easy for students and would spark their interest in the subject.

Extramarks helps students improve their skills and polish their problem-solving skills. Apart from the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, students will be able to access other study materials like live classes, past years’ question papers, sample papers, complete syllabus etc.

Mathematics may seem like a major challenge with its vast academic curriculum prescribed for Class 11 but with the aid of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, students would be more confident with their preparation for this subject. Students learn a lot from the calculations provided as a part of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. If students try to adhere to the format of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, they will excel in Mathematics and in every other subject they are studying. Students can improve their academic performance by focusing on Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. These solutions can be downloaded from the Extramarks website or mobile application. NCERT books have made it clear which topic will have how much weightage and this pattern has been thoroughly followed by the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 comprise comprehensive solutions to questions and all the formulae used have been highlighted for the convenience of students. Students who thoroughly study the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will be familiar with the important questions and efficient ways to solve these questions in order to save time in exams.These Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 contain all of the important tricks and solutions that will aid them in their upcoming Class 12 and competitive examinations such as the Joint Entrance Exam (JEE).

Exercise 9.3

This Exercise- 9.3 Class 11th questions have been solved in a descriptive manner. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 comprise solutions to all the 32 questions of the given chapter- Sequences and Series. There are a total of 4 exercises and a miscellaneous exercise in Chapter 9- Sequences and Series. Solutions to all these questions have been provided by Extramarks. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are focused on the holistic development of students. Students are advised to continuously practise the important questions that might constitute in Class 11 annual examination. All the important questions and notes have been provided within the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 to facilitate self-guided learning among students. Students can also do a self-assessment of their preparation for the upcoming examinations with the help of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3.

The academic curriculum of Mathematics for Class 11 is very different from that of Class 10. This is because the curriculum of Class 10 merely comprises a fundamental syllabus of basic Mathematics while the syllabus of Class 11 is a much broader version. Accordingly, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are available to assist students during this transition. To enable students to be appropriately prepared for the increase in the academic curriculum in Class 11, Extramarks is providing all kinds of solutions pertaining to the assessments provided as a part of the NCERT curriculum. As a result, students will be able to focus on their studies when they have access to Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3.Therefore, students must try to learn and understand the NCERT book questions and then take help from the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. Students can use resources like past years’ question papers and sample question papers provided by Extramarks to evaluate themselves. Constant revision and continuous practise are indispensable to scoring well in annual examinations. Students must revise and practice with the help of Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. Students must stay updated with new question patterns and problem-solving methods with the help of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would also help students prepare themselves for Multiple Choice Questions (MCQs) as well.

NCERT Solutions for Class 11 Maths Chapters

Extramarks provides material for all boards: CBSE, ICSE, and other State Boards like Rajasthan Board, Haryana Board etc.Extramarks, as a reputable online learning platform, ensures that all solutions available to students are both precise and comprehensive. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are very helpful to students. Now Extramarks has made available solutions for the entire academic curriculum of Mathematics prescribed for Class 11 by the NCERT. As a result, students can begin learning each chapter without hesitation because solutions have already been made available on the Extramarks online learning portal.The main aim of Extramarks is to make students reliant on themselves by guiding them toward the right path in learning. Students must try to follow a strict timetable for learning so that they are able to comprehensively study the entire syllabus in time for the examinations. In this regard, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have made the solutions of Exercise 9.3 available for students. Compliance with a proper routine in a student’s life will help them complete every chapter before examinations. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 adhere to the pattern provided within the prescribed NCERT textbooks

Accuracy is the primary objective of the Extramarks learning portal. This makes it self-evident that students can rely on the solutions and formats of problem-solving provided by the learning resources available on the Extramarks website. Students should focus on learning all academic disciplines in detail. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help students to qualify for all examinations. Students must know that if they require any assistance in their study and they feel doubtful about specific questions, they must take help from Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. The solutions for all the exercises that have been provided alongside the chapters of the prescribed textbook of Mathematics for Class 11 have been compiled at the Extramarks online portal for learning. Students must go through each chapter’s solutions given by Extramarks. They are very helpful as Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3.

The solutions to each chapter of the prescribed textbook of Mathematics for Class 11 have been engineered by Extramarks. Students can access these solutions easily through the Extramarks website as well as through the Extramarks mobile application. The entire curriculum of Mathematics for Class 11 has been divided into six units, which have been further subdivided into sixteen chapters. The following are the chapters:

NCERT Solutions Class 11 Chapter 1- Sets

NCERT Solutions Class 11 Chapter 2 – Relations and Functions

NCERT Solutions Class 11 Chapter 3- Trigonometric Functions

NCERT Solutions Class 11 Chapter 4- Principle of Mathematical Induction

NCERT Solutions Class 11 Chapter 5- Complex Numbers and Quadratic Equations

NCERT Solutions Class 11 Chapter 6- Linear Inequalities

NCERT Solutions Class 11 Chapter 7- Permutations and Combinations

NCERT Solutions Class 11 Chapter 8- Binomial Theorem

NCERT Solutions Class 11 Chapter 9- Sequences and Series

NCERT Solutions Class 11 Chapter 10- Straight Lines

NCERT Solutions Class 11 Chapter 11- Conic Sections

NCERT Solutions Class 11 Chapter 12- Introduction to Three Dimensional Geometry

NCERT Solutions Class 11 Chapter 13- Limits and Derivatives

NCERT Solutions Class 11 Chapter 14- Mathematical Reasoning

NCERT Solutions Class 11 Chapter 15- Statistics

NCERT Solutions Class 11 Chapter 16- Probability

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

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1

NCERT Solution Class 11 Maths of Chapter 9 Exercises

The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are extremely useful when students are faced with assessments that appear to be more complex than the ones they are used to.. Chapter 9 consists of 4 exercises and an additional exercise consisting of miscellaneous questions. In Exercise 9.1, there are a total of 14 questions, then in Exercise 9.2, there are a total of 18 questions, and the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 is made up of an aggregate of 32 questions. Further, in Exercise 9.4, there are a total of 10 questions. Finally, in the Miscellaneous Exercises, there are a total of 32 questions. Each of these questions has been solved with unparalleled precision and expertise. Undoubtedly, students can trust the Extramarks application.

In Mathematics, a tabular form of representing information is very important. This form of disseminating information helps students grasp information in a better and more simplified manner. Students must understand the importance of a tabular form of organising information and the ways in which it can be used. In the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, a table of exercises has been included in order to aid students in better understanding the scheme of mark distribution relevant to the exerciseThe Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help students figure out the appropriate method to respond to a problem designated to them. The academic curriculum of Class 11 is designed in a way that prepares students for the board examinations of Class 12 and other competitive examinations. Therefore, students must be sincere in their efforts at learning the academic curriculum of Class 11. Students should devise effective time management plans and techniques while studying for Class 11 Mathematics.By studying the solutions prepared in tandem with knowledgeable mathematics mentors, students will gain a better understanding of fundamental concepts, which will expand the horizons of their knowledge. Problem-solving is an important strategy for dealing with mathematical concepts.

Students are highlyrecommended to go through the table provided below in order to enlighten themselves with regard to the pattern of questions and the concurrent schema of marks distribution associated with the chapter on Sequences and Series.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3

The NCERT Solutions provided by Extramarks have been very helpful to students. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will make all the students’ doubts clear. The mathematics curriculum for Class 11 covers a variety of significant topics that are important components of the overall higher-level study of mathematics. Many concepts introduced in the mathematics curriculum in Class 11 are not only critical for the Class 11 examinations but are also significant for the preparation for various entrance examinations. Class 11 Maths NCERT Solutions Chapter 9, Exercise 9.3 should be used by students preparing for the Class 11 Mathematics examination.StUsing Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3, students can understand all of the designated questions.Students musbecomeet familiar with all of the working principles of Mathematics for Class 11 because they will be evaluated on the basis of their knowledge of these principles also in Class 12.

Every chapter in Class 11 is significant for competitive examinations, so overlooking any chapter would not be a wise choice for students. In order to adequately complete all the chapters, topic-wise Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 will help students in achieving good marks. Class 11 is the foundation stone for the academic future of students. Therefore, students must keep in mind that only a strong foundation can lead them towards a bright future.

Extramarks is a holistic solution for all the concerns and needs of students pertaining to the academic curriculum. Students must secure access to Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. Students can procure the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 by visiting the Extramarks website or by downloading the mobile application, which contains these solutions in PDF format. Even if students do not have Internet access, the PDF format of such solutions can be helpful. Extramarks has made learning easier for students by ensuring the availability of easy-to-understand solutions and short notes.

To resolve the exercise questions which are a part of the curriculum of Mathematics in Class 11, all of the important topics must be thoroughly understood. Chapter 9 is titled Sequences and Series. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are a prerequisite for better results in Mathematics. Sequences and Series is an important theme in the NCERT textbook which must be read by students daily and in-depth. When it comes to competitive examinations, Sequences and Series are extremely important. Students must complete the exercises in Chapter 9 after reviewing the topics. The topics which comprise the syllabus of Mathematics in Class 11 are broad but simple, and a deeper understanding of these topics will help students to solve questions in the annual Mathematics examination.

To prepare for the Mathematics examination, students can use the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. The quality of examination preparation can be greatly improved by using online learning resources. To effectively prepare for the annual examinations, study materials from reliable sources are required. Students can access all of the useful resources through the Extramarks website and mobile application.

Topics Covered in Exercise 9.3 of Chapter 9 Sequences and Series

re several important topics in Exercise 9.3 of Chapter 9- Sequences and Series. The chapter is mainly based on Geometric Progression (GM) and the relationship between Arithmetic Mean and Geometric Progression (AM and GM). Therefore, students must know the important topics of the chapter. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been carefully included on the Extramarks learning portal, with careful inclusion of all the important topics.These solutions have been modelled in a pattern which will guide students in self-study and in identifying any possible concerns with regard to the curriculum. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 would supplement the digital education programme and smart classroom infrastructure.

The solutions have been crafted with reference to the contributions made by proficient mentors of Mathematics on the Extramarks learning platform. Students will excel in the Class 11 examinations if they will follow the pattern of problem-solving provided within the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 designed by Extramarks. Students must check the solutions of Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 so that they will learn how to get marks by writing the correct steps. All the topics of Chapter 9- Sequences and Series have been covered in Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3. That is why students must go through the solutions of all the exercises to familiarise themselves with the different important topics in Chapter 9. To learn about other chapters as well, students can browse through the Extramarks application, as everything is available there.

Some topics that have been covered by Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are Arithmetic Progression, Geometric Progression, Geometric Mean, Arithmetic Mean and the Relationship between the Arithmetic and Geometric Progression.Extramarks’ Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 also covers another important topic involving the sum to n terms of special series.Important Formulas to Solve Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3

The NCERT Class 11 Chapter 9 covers a lot of concepts which are a part of the theme of Algebra. The solutions provided as a part of the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are equipped with appropriate formulas and will help students learn to adequately apply them.

The important formulas have been covered by the provided solutions by Extramarks and they have been provided in a framework which would enhance the conceptual understanding of students. Students will find it easier to remember formulae that have been frequently used in the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 if they practise on a daily basis.The important formulas to solve Class 11 Maths Chapter 9 Exercise 9.3 have been curated by Extramarks to ensure error-free and quality content.

To practise the Mathematics exercise questions for Class 11, all of the important topics must be thoroughly understood. The Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are essential for achieving perfection in Mathematics. Sequence and Series concepts from the NCERT textbook should be read by students on a regular basis. Students must complete the exercises in Chapter 9 after reviewing the topics. Class 11 Maths topics are broad but simple, and a deeper understanding of the topics is required in order to answer questions in the annual Mathematics examination.

Q.1 Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.

Ans.

$\begin{array}{l} Common ratio of given G.P. = 2 \\ 8^{th} term of A.P. = 192 \\ {ar}^{8 - 1} = 192 \\ a . 2^{7} = 192 \\ a = \frac{192}{2^{7}} = 1 .5 \\ 12^{th} term of A.P. = {ar}^{12 - 1} \end{array}$ $\begin{array}{l} = 1 .5 {(2)}^{11} \\ = 1 .5 \times 2048 \\ = 3072 \end{array}$

Q.2 The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Ans.

$\begin{array}{l} Let first term of G.P. be a and common ratio be r. \\ 5^{th} term of G.P. = p \\ \Rightarrow T_{5} = p \\ \Rightarrow p = {ar}^{4} \\ 8^{th} term of G.P. = q \\ \Rightarrow T_{8} = q \\ \Rightarrow q = {ar}^{7} \\ 11^{th} term of G.P. = s \\ \Rightarrow T_{11} = s \\ \Rightarrow s = {ar}^{10} \\ L.H.S. = q^{2} \\ = {({ar}^{7})}^{2} \\ = a^{2} r^{14} \\ R.H.S.= ps \\ = ({ar}^{4}) ({ar}^{10}) \\ = a^{2} r^{14} \\ So, L.H.S.=R.H.S. \\ Hence proved. \end{array}$

Q.3 The 4^th term of a G.P. is square of its second term, and the first term is –3. Determine its 7^th term.

Ans.

$\begin{array}{l} First term of G.P. = - 3 \\ Let common ratio of G.P. = r \\ According to given condition, \\ T_{4} = {(T_{2})}^{2} \\ \Rightarrow {ar}^{3} = {(ar)}^{2} \\ \Rightarrow (- 3) r^{3} = {(- 3)}^{2} r^{2} \\ \Rightarrow r = - 3 \\ Then, T_{7} = {ar}^{6} \\ = (- 3) {(- 3)}^{6} \\ = {(- 3)}^{7} \\ = - 2187 \\ Thus, 7^{th} term of G.P. is - 2187 . \end{array}$

Q.4

$\begin{array}{l} Which term of the following sequences: \\ (a) . 2, 2 \sqrt{2}, 4, . .. is 128? \\ (b) . 3, 3 \sqrt{3}, 3, . .. is 729? \\ (c) . \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, . .. is \frac{1}{19683} ? \end{array}$

Ans.

$\begin{array}{l} (a) Let t_{n} = 128, a = 2, \\ r = \frac{2 \sqrt{2}}{2} = \sqrt{2} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} t_{n} = {ar}^{n - 1} \\ 128 = 2 {(\sqrt{2})}^{n - 1} \\ \Rightarrow 2^{7} = {(2)}^{1 + \frac{n - 1}{2}} \\ \Rightarrow 7 = 1 + \frac{n - 1}{2} \\ \Rightarrow 6 \times 2 = n - 1 \Rightarrow n = 13 \\ Thus, {128 is 13}^{th} term of G.P. \\ (b) Let t_{n} = 729, a = \sqrt{3}, \\ r = \frac{3}{\sqrt{3}} = \sqrt{3} \\ t_{n} = {ar}^{n - 1} \\ 729 = \sqrt{3} {(\sqrt{3})}^{n - 1} \\ \Rightarrow 3^{6} = {(3)}^{\frac{1}{2} + \frac{n - 1}{2}} \\ \Rightarrow 6 = \frac{1}{2} + \frac{n - 1}{2} \\ \Rightarrow 6 \times 2 = 1 + n - 1 \\ \Rightarrow n = 12 \\ Thus, 729 {is 12}^{th} term of G.P. \\ (c) Let t_{n} = 729, a = \frac{1}{3}, \\ r = \frac{(\frac{1}{9})}{(\frac{1}{3})} = \frac{1}{3} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} t_{n} = {ar}^{n - 1} \\ \frac{1}{19683} = \frac{1}{3} {(\frac{1}{3})}^{n - 1} \\ \Rightarrow {(\frac{1}{3})}^{9} = {(\frac{1}{3})}^{n} \\ \Rightarrow 9 = n \\ \Rightarrow n = 9 \\ Thus, \frac{1}{19683} {is 9}^{th} term of G.P. \end{array}$

Q.5

For what values of x, the numbers - \frac{2}{7},x, - \frac{7}{2} are in G.P.?

Ans.

$\begin{array}{l} - \frac{2}{7},x, - \frac{7}{2} are in G.P., then \\ Common ratio = \frac{x}{- (\frac{2}{7})} \\ = - \frac{7 x}{2} \\ Also, common ratio = \frac{(- \frac{7}{2})}{x} \\ = - \frac{7}{2 x} \\ Then, \\ - \frac{7 x}{2} = - \frac{7}{2 x} \\ \Rightarrow x^{2} = 1 \\ \Rightarrow x = \pm 1 \end{array}$

Q.6

$\begin{array}{l} Find the sum to indicated number of terms in each of the geometric \\ progressions given below. \\ i. 0.15, 0.015, 0.0015,… 20 terms. \\ ii. \sqrt{7}, \sqrt{21}, 3 \sqrt{7}, . .. n terms. \\ iii. 1, - {a, a}^{2}, - a^{3}, . .. n terms (if a \neq - 1). \\ {iv. x}^{3} {, x}^{5} {, x}^{7}, . .. n terms (if x \neq ± 1). \end{array}$

Ans.

$\begin{array}{l} i . \\ 0.15, 0 .0 15, 0 .00 15, \dots 20 terms \\ Since, S_{n} = \frac{a (1 - r^{n})}{1 - r} \\ a = 0 .15, \\ r = \frac{0 .015}{0 .15} = 0 .1 \\ S_{n} = \frac{0 .15 {1 - {(0 .1)}^{20}}}{1 - 0 .1} \\ = \frac{0 .15 {1 - {(0 .1)}^{20}}}{0 .9} \\ = \frac{15}{90} {1 - {(0 .1)}^{20}} \\ = \frac{1}{6} {1 - {(0 .1)}^{20}} \end{array}$

$\begin{array}{l} i i . \\ \sqrt{7}, \sqrt{21}, 3 \sqrt{7}, . .. n terms \\ Since, S_{n} = \frac{a (r^{n} - 1)}{r - 1} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} a = \sqrt{7}, \\ r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \end{array}$

$\begin{array}{l} S_{n} = \frac{\sqrt{7} {{(\sqrt{3})}^{n} - 1}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ = \frac{\sqrt{7} {{(\sqrt{3})}^{n} - 1}}{3 - 1} \times (\sqrt{3} + 1) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} iii . \end{array}$

$\begin{array}{l} 1, - a, a^{2}, - a^{3}, . .. nterms \\ First term of G.P. = 1 \\ Common ratio of G.P. = \frac{- a}{1} \\ = - a \\ Since, S_{n} = \frac{a (1 - r^{n})}{1 - r} \\ S_{n} = \frac{1 {1 - {(- a)}^{n}}}{1 + a} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} iv . \\ x^{3}, x^{5}, x^{7}, . .. n terms \\ First term of G.P. = x^{3} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} Common ratio of G.P. = \frac{x^{5}}{x^{3}} \\ = x^{2} \\ Since, S_{n} = \frac{a (r^{n} - 1)}{r - 1} \\ S_{n} = \frac{x^{3} {1 - {(x^{2})}^{n}}}{1 - x^{2}} \\ S_{n} = \frac{x^{3} (1 - x^{2 n})}{1 - x^{2}} \end{array}$

Q.7

Evaluate : \sum_{k = 1}^{11} (2 + 3^{k}) .

Ans.

$\begin{array}{l} We have, \\ \sum_{k = 1}^{11} (2 + 3^{k}) = (2 + 3^{1}) + (2 + 3^{2}) + (2 + 3^{3}) + . .. + (2 + 3^{11}) \\ = 2 \times 11 + (3^{1} + 3^{2} + 3^{3} + . .. + 3^{11}) \\ = 22 + \frac{3 (3^{11} - 1)}{3 - 1} \\ = 22 + \frac{3}{2} (3^{11} - 1) \end{array}$

Q.8

$\begin{array}{l} The sum of first three terms of a G.P. is \frac{39}{10} and their product is 1. \\ Find the common ratio and the terms. \end{array}$

Ans.

$\begin{array}{l} Let three terms in A.P. are \frac{a}{r}, a, ar . \\ Then according to given conditions: \\ \frac{a}{r} + a + ar = \frac{39}{10} \\ a (\frac{1}{r} + 1 + r) = \frac{39}{10} . .. (i) \\ and \frac{a}{r} \times a \times ar = 1 \\ a^{3} = 1 \\ \Rightarrow a = 1 \\ Substituting value of a in equation (i), we get \\ 1 (\frac{1}{r} + 1 + r) = \frac{39}{10} \\ \Rightarrow (\frac{1 + r + r^{2}}{r}) = \frac{39}{10} \\ 10 (r^{2} + r + 1) = 39 r \\ 10 r^{2} + 10 r + 10 = 39 r \\ 10 r^{2} - 29 r + 10 = 0 \\ 10 r^{2} - 25 r - 4 r + 10 = 0 \\ 5 r (2 r - 5) - 2 (2 r - 5) = 0 \\ (2 r - 5) (2 r - 5) = 0 \\ \Rightarrow r = \frac{5}{2}, \frac{5}{2} \end{array}$

$\begin{array}{l} Thus, first term of G.P. is 1 and common difference is \frac{5}{2} . \\ Three terms are: \frac{1}{\frac{5}{2}}, 1, 1 \times \frac{5}{2} \Rightarrow \frac{2}{5}, 1, \frac{5}{2} \\ Or \frac{1}{\frac{2}{5}}, 1, 1 \times \frac{2}{5} \Rightarrow \frac{5}{2}, 1, \frac{2}{5} \end{array}$

Q.9 How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

Ans.

$\begin{array}{l} G . P . is 3, 3^{2}, 3^{3}, \dots \\ a = 3, r = \frac{3^{2}}{3} = 3 {and S}_{n} = 120 \\ S_{n} = a \frac{(r^{n} - 1)}{(r - 1)} \end{array}$

$\begin{array}{l} 120 = 3 (\frac{3^{n} - 1}{3 - 1}) \\ 120 = \frac{3}{2} (3^{n} - 1) \\ 120 \times \frac{2}{3} = (3^{n} - 1) \\ 81 = 3^{n} \\ \Rightarrow 3^{4} = 3^{n} \\ \Rightarrow n = 4 \\ Thus, 4 terms of G.P. are needed to give the sum 120. \end{array}$

Q.10 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the first term, the common ratio and the sum to n terms of the G.P.

Ans.

$\begin{array}{l} Let first term of G.P. is a and common ratio is r. Then, \\ {six terms of G.P. are: a, ar, ar}^{2} {, ar}^{3} {, ar}^{4} {, ar}^{5} {, ar}^{6} . \\ According to first condition: \\ a + ar + {ar}^{2} = 16 \\ \Rightarrow a (1 + r + r^{2}) = 16 . .. (i) \\ {ar}^{3} + {ar}^{4} + {ar}^{5} = 128 \\ \Rightarrow {ar}^{3} (1 + r + r^{2}) = 128 . .. (ii) \\ Dividing equation (ii) by equation (i), we get \\ \frac{{ar}^{3} (1 + r + r^{2})}{a (1 + r + r^{2})} = \frac{128}{16} \\ \Rightarrow r^{3} = 8 \\ \Rightarrow r = 2 \\ Putting r = 2 in equation (i), we get \\ a (1 + 2 + 2^{2}) = 16 \\ a = \frac{16}{7} and r = 2 \\ Sum of n terms is: \\ S_{n} = \frac{a (r^{n} - 1)}{(r - 1)} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{\frac{16}{7} (2^{n} - 1)}{(2 - 1)} \\ = \frac{16}{7} (2^{n} - 1) \end{array}$

Q.11 Given a G.P. with a = 729 and 7^th term 64, determine S₇.

Ans.

$\begin{array}{l} First term of G.P. = (a) \\ = 729 \\ 7^{th} term of G.P. = T_{7} \\ = 64 \\ Then, {ar}^{6} = 64 \\ \Rightarrow 729 r^{6} = 64 \\ r^{6} = \frac{64}{729} \\ r^{6} = \frac{2^{6}}{3^{6}} \\ \Rightarrow r = \frac{2}{3} \\ S_{n} = \frac{a (1 - r^{n})}{(1 - r)} \\ \Rightarrow S_{7} = \frac{729 (1 - \frac{2^{n}}{3^{n}})}{(1 - \frac{2}{3})} \end{array}$

$\begin{array}{l} \Rightarrow S_{7} = 729 \times 3 (1 - \frac{2^{7}}{3^{7}}) \\ \Rightarrow S_{7} = 729 \times 3 (1 - \frac{128}{2187}) \\ \Rightarrow S_{7} = 2187 (\frac{2187 - 128}{2187}) \\ \Rightarrow S_{7} = 2059 \\ Thus, sum of 7 terms of G.P. is 2059. \end{array}$

Q.12 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Ans.

$\begin{array}{l} Let G.P. be: \\ {a, ar, ar}^{2} {, ar}^{3} {, ar}^{4} {, ar}^{5}, . .. \\ Then according to first condition, \\ a + ar = - 4 \\ or a (1 + r) = - 4 \\ or a = \frac{- 4}{1 + r} . .. (i) \\ Then according to second condition, \\ T_{5} = {4T}_{3} \\ \Rightarrow {ar}^{4} = 4 ({ar}^{2}) \\ \Rightarrow r^{2} = 4 \\ \Rightarrow r = \pm 2 \\ Putting r = 2 in equation (i), we get \end{array}$ $\begin{array}{l} a = \frac{- 4}{1 + 2} \\ = - \frac{4}{3} \end{array}$

$\begin{array}{l} Putting r = 2 in equation (i), we get \\ a = \frac{- 4}{1 - 2} \\ = \frac{- 4}{- 1} \\ = 4 \\ Therefore, G.P. when a = - \frac{4}{3} and r = 2 \\ - \frac{4}{3}, - \frac{4}{3} (2), - \frac{4}{3} {(2)}^{2}, - \frac{4}{3} {(2)}^{3}, - \frac{4}{3} {(2)}^{4}, - \frac{4}{3} {(2)}^{5}, . .. \\ or - \frac{4}{3}, - \frac{8}{3}, - \frac{16}{3}, - \frac{32}{3}, - \frac{64}{3}, - \frac{128}{3}, . .. \\ Therefore, G.P. when a = 4 and r = - 2 \\ 4, 4 (- 2), 4 {(- 2)}^{2}, 4 {(- 2)}^{3}, 4 {(- 2)}^{4}, 4 {(- 2)}^{5}, . .. \\ or 4, - 8, 16, - 32, . .. \\ Therefore, the required G.P. is \\ - \frac{4}{3}, - \frac{8}{3}, - \frac{16}{3}, - \frac{32}{3}, - \frac{64}{3}, - \frac{128}{3}, . .. \\ or 4, - 8, 16, - 32, . .. \end{array}$

Q.13 If the 4^th, 10^th and 16^th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Ans.

$Let first term of G.P. be ‘a’ and common ratio be ‘r’.$ $\begin{array}{l} T_{4} = x \\ \Rightarrow {ar}^{3} = x \\ T_{10} = y \\ \Rightarrow {ar}^{9} = y \\ T_{16} = z \\ \Rightarrow {ar}^{15} = z \\ Now, \frac{y}{x} = \frac{{ar}^{9}}{{ar}^{3}} \\ = r^{6} \\ and \frac{z}{y} = \frac{{ar}^{15}}{{ar}^{9}} \\ = r^{6} \\ \Rightarrow \frac{y}{x} = \frac{z}{y} \\ \Rightarrow x, y, z are in G.P. \end{array}$

Q.14 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Ans.

$\begin{array}{l} Let S_{n} = 8 + 88 + 888 + 8888 + \dots \\ = 8 (1 + 11 + 111 + 1111 + . ..) \\ = \frac{8}{9} (9 + 99 + 999 + 9999 + . ..) \\ = \frac{8}{9} {(10 - 1) + (10^{2} - 1) + (10^{3} - 1) + (10^{4} - 1) + . ..} \\ = \frac{8}{9} {10 + 10^{2} + 10^{3} + 10^{4} + . .. - 1 - 1 - 1 - 1 - . .. n times} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{8}{9} {\frac{10 (10^{n} - 1)}{10 - 1} - n} [∵ S_{n} = \frac{a (r^{n} - 1)}{r - 1}] \\ = \frac{8}{9} {\frac{10 (10^{n} - 1)}{9} - n} \\ = \frac{80}{81} (10^{n} - 1) - \frac{8}{9} n \end{array}$

Q.15

$\begin{array}{l} Find the sum of the products of the corresponding terms \\ of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2, \frac{1}{2} . \end{array}$

Ans.

$\begin{array}{l} Sum of the products of the corresponding terms of the sequences \\ 2, 4, 8, 16, 32 and 128, 32, 8, 2, \frac{1}{2} \\ = 2 \times 128 + 4 \times 32 + 8 \times 8 + 16 \times 2 + 32 \times \frac{1}{2} \\ = 256 + 128 + 64 + 32 + 16 \\ = \frac{256 \{1 - {(\frac{1}{2})}^{5}\}}{1 - \frac{1}{2}} = \frac{256 (\frac{32 - 1}{32})}{\frac{1}{2}} \\ = 256 \times 2 \times \frac{31}{32} = 496 \end{array}$

Q.16 Show that the products of the corresponding terms of the sequences a, ar, ar², … arⁿ^{– 1}and A, AR, AR², … AR^n–1 form a G.P, and find the common ratio.

Ans.

$\begin{array}{l} The products of the corresponding terms of the sequences \\ a, ar, {ar}^{2}, \dots {ar}^{n}^{- 1} and A, AR, {AR}^{2}, \dots {AR}^{n - 1} \\ = aA, aARr, {aAR}^{2} r^{2}, \dots {,aAR}^{n}^{- 1} {r^{n}}^{- 1} \\ Common ratio = \frac{aARr}{aA} = Rr \\ Common ratio = \frac{{aAR}^{2} r^{2}}{aARr} = Rr \\ Since, common ratio is same i.e., rR. \\ So, aA, aARr, {aAR}^{2} r^{2}, \dots {,aAR}^{n}^{- 1} {r^{n}}^{- 1} is in G.P. \end{array}$

Q.17 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

Ans.

$\begin{array}{l} {Let four numbers of G.P. are a, ar, ar}^{2} {, ar}^{3} . \\ According to first condition, \\ T_{3} - T_{1} = 9 \\ \Rightarrow {ar}^{2} - a = 9 \\ \Rightarrow a (r^{2} - 1) = 9 . .. (i) \\ According to second condition, \\ T_{2} - T_{4} = 18 \\ \Rightarrow ar - {ar}^{3} = 18 \\ \Rightarrow ar (1 - r^{2}) = 18 \\ \Rightarrow ar (r^{2} - 1) = - 18 . .. (ii) \\ Dividing equation (ii) by equation (i), we get \\ \frac{ar (r^{2} - 1)}{a (r^{2} - 1)} = - \frac{18}{9} \\ \Rightarrow r = - 2 \\ Putting r= - 2, in equation (i), we get \\ a \{{(- 2)}^{2} - 1\} = 9 \\ a = \frac{9}{3} \\ = 3 \\ Thus, the four terms of G.P. are: \\ 3, 3 (- 2), 3 {(- 2)}^{2}, 3 {(- 2)}^{3} \\ i . e ., 3, - 6, 12, - 24 \end{array}$

Q.18

$\begin{array}{l} {If the p}^{th} {, q}^{th} {and r}^{th} terms of a G.P. are a and b, respectively. \\ {Prove that a}^{q - r} b^{r - p} c^{p - q} =1. \end{array}$

Ans.

$\begin{array}{l} Let first term of G.P. is A and common ratio is R. \\ Given that: \\ T_{p} = a \Rightarrow {AR}^{p - 1} = a \\ T_{q} = b \Rightarrow {AR}^{q - 1} = b \\ T_{r} = c \Rightarrow {AR}^{r - 1} = c \\ L . H . S . = a^{q - r} b^{r - p} c^{p - q} \\ = {({AR}^{p - 1})}^{q - r} {({AR}^{q - 1})}^{r - p} {({AR}^{r - 1})}^{p - q} \\ = A^{q - r + r - p + p - q} R^{(p - 1) (q - r) + (q - 1) (r - p) (r - 1) (p - q)} \\ = A^{0} R^{pq - pr - q + r + qr - qp - r + p + rp - rq - p + q} \\ = A^{0} R^{0} \\ = 1 \times 1 \\ = 1 = R . H . S . \end{array}$

Q.19 If the first and the n^th term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.

Ans.

$\begin{array}{l} First term of G.P. = a \\ n^{th} term of G.P. = b \\ Let common ratio = r \\ The G.P. is \\ {a, ar, ar}^{2} {, ar}^{3} {, ar}^{4}, . .., b \\ and b = {ar}^{n-1} \\ Product of n terms = a \times ar \times {ar}^{2} \times {ar}^{3} \times {ar}^{4} \times . .. \times b \\ = a \times ar \times {ar}^{2} \times {ar}^{3} \times {ar}^{4} \times . .. \times {ar}^{n-1} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = a^{n} \times r^{1 + 2 + 3 + 4 + . .. + (n - 1)} \\ P = a^{n} \times r^{\frac{(n - 1) (n - 1 + 1)}{2}} \\ P = a^{n} \times r^{\frac{n (n - 1)}{2}} \\ Squarring both sides, we get \\ P^{2} = a^{2 n} r^{n (n - 1)} \\ = {(a^{2} r^{n - 1})}^{n} \\ = {(a \times {ar}^{n - 1})}^{n} \\ P^{2} = {(ab)}^{n} [∵ b = {ar}^{n - 1}] \\ Hence Proved. \end{array}$

Q.20

$\begin{array}{l} Show that the ratio of the sum of first n terms of a G.P. \\ to the sum of terms from {(n+1)}^{th} to {(2n)}^{th} term is \frac{1}{r^{n}} . \end{array}$

Ans.

$\begin{array}{l} Let first term of G.P. is a and common ratio is r. Then, \\ Sum of first n terms (S_{n}) = \frac{a (r^{n} -1)}{(r-1)} \\ Sum of 2n terms (S_{2n}) = \frac{a (r^{2n} -1)}{(r-1)} \\ Sum of terms from {(n+1)}^{th} to {(2n)}^{th} term \\ {= S}_{2n} {-S}_{n} \end{array}$

$\begin{array}{l} = \frac{a (r^{2n} -1)}{(r-1)} – \frac{a (r^{n} -1)}{(r-1)} \\ = \frac{a}{(r-1)} (r^{2n} {-1-r}^{n} +1) \\ = \frac{{ar}^{n}}{(r-1)} (r^{n} -1) \\ The required ratio = \frac{S_{n}}{S_{2n} {-S}_{n}} \\ = \frac{\frac{a (r^{n} -1)}{(r-1)}}{\frac{{ar}^{n}}{(r-1)} (r^{n} -1)} \\ = \frac{1}{r^{n}} \\ Thus, the ratio of the sum of first n terms of a G.P. to \\ the sum of terms from {(n+1)}^{th} to {(2n)}^{th} term is \frac{1}{r^{n} .} \end{array}$

Q.21 If a, b, c and d are in G.P. show that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)².

Ans.

$\begin{array}{l} Since, a, b, c, d are in G.P. \\ So, \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r (let) \\ \Rightarrow b = ar, c = br = {ar}^{2} \\ and d = cr = {ar}^{4} \\ L . H . S . = (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = (a^{2} + a^{2} r^{2} + a^{2} r^{4}) (a^{2} r^{2} + a^{2} r^{4} + a^{2} r^{6}) \\ = a^{2} (1 + r^{2} + r^{4}) \times a^{2} r^{2} (1 + r^{2} + r^{4}) \\ = a^{4} r^{2} {(1 + r^{2} + r^{4})}^{2} \\ R . H . S . = {(ab + bc + cd)}^{2} \\ = {(a \times ar + ar \times {ar}^{2} + {ar}^{2} \times {ar}^{4})}^{2} \\ = a^{4} r^{2} {(1 + r^{2} + r^{4})}^{2} \\ L . H . S . = R . H . S . \\ Therefore, \\ (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = {(ab + bc + cd)}^{2} \\ Hence proved. \end{array}$

Q.22 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Ans.

$\begin{array}{l} Let two numbers between 3 and 81 are x and y in such a way \\ that 3, x, y, 81 are in G.P. \\ So, first term of G.P. = 3 \\ Common difference = r \\ 4^{th} term of G.P. = {ar}^{3} \\ \Rightarrow 3 r^{3} = 81 \\ \Rightarrow r^{3} = \frac{81}{3} \\ \Rightarrow r = \sqrt[3]{27} \\ = 3 \end{array}$

$\begin{array}{l} ∴ x = T_{2} \\ = ar \\ = 3 \times 3 \\ = 9 \\ andy = T_{3} \\ = {ar}^{2} \\ = 3 \times 3^{2} \\ = 27 \\ Thus, two numbers between 3 and 81 are 9 and 27. \end{array}$

Q.23

Find the value of n so that \frac{a^{n+1} {+b}^{n+1}}{a^{n} {+b}^{n}} may be the geometric mean between a and b.

Ans.

$\begin{array}{l} Geometric mean between a and b = \sqrt{ab} \\ \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} = \sqrt{ab} \\ \Rightarrow {(ab)}^{\frac{1}{2}} (a^{n} + b^{n}) = a^{n + 1} + b^{n + 1} \\ \Rightarrow a^{n + \frac{1}{2}} b^{\frac{1}{2}} + a^{\frac{1}{2}} b^{n + \frac{1}{2}} = a^{n + 1} + b^{n + 1} \\ \Rightarrow a^{n + \frac{1}{2}} b^{\frac{1}{2}} - a^{n + 1} = b^{n + 1} - a^{\frac{1}{2}} b^{n + \frac{1}{2}} \\ \Rightarrow a^{n + \frac{1}{2}} (b^{\frac{1}{2}} - a^{\frac{1}{2}}) = b^{n + \frac{1}{2}} (b^{\frac{1}{2}} - a^{\frac{1}{2}}) \\ \Rightarrow a^{n + \frac{1}{2}} = b^{n + \frac{1}{2}} \\ \Rightarrow {(\frac{a}{b})}^{n + \frac{1}{2}} = 1 \\ \Rightarrow {(\frac{a}{b})}^{n + \frac{1}{2}} = {(\frac{b}{a})}^{0} \\ \Rightarrow n + \frac{1}{2} = 0 \\ \Rightarrow n = - \frac{1}{2} \end{array}$

Q.24

$\begin{array}{l} The sum of two numbers is 6 times their geometric mean, show that \\ numbers are in the ratio (3+2 \sqrt{2}) : (3 - 2 \sqrt{2}) . \end{array}$

Ans.

$\begin{array}{l} Let two numbers be a and b. \\ G.M. of a and b = \sqrt{ab} \\ Then, according to given condition, \\ a + b = 6 \sqrt{ab} . .. (i) \\ ∵ {(a - b)}^{2} = {(a + b)}^{2} - 4 ab \\ \Rightarrow {(a - b)}^{2} = {(6 \sqrt{ab})}^{2} - 4 ab \\ = 36 ab - 4 ab \\ = 32 ab \\ \Rightarrow a - b = \sqrt{32 ab} \\ \Rightarrow a - b = 4 \sqrt{2} \sqrt{ab} . .. (ii) \end{array}$

$\begin{array}{l} Solving equation (i) and equation (ii), we get \\ a = (3 + 2 \sqrt{2}) \sqrt{ab} and b = (3 - 2 \sqrt{2}) \sqrt{ab} \\ Then, \\ ratio of the numbers = \frac{(3 + 2 \sqrt{2}) \sqrt{ab}}{(3 - 2 \sqrt{2}) \sqrt{ab}} \\ = (3 + 2 \sqrt{2}) : (3 - 2 \sqrt{2}) \end{array}$

Q.25

$\begin{array}{l} If A and G be A.M. and G.M., respectively between two positive numbers, \\ prove that the numbers are A± \sqrt{(A+G) (A - G)} . \end{array}$

Ans.

$\begin{array}{l} Let two numbers are a and b. \\ Then, A.M. = \frac{a + b}{2} \\ \Rightarrow A = \frac{a + b}{2} \\ \Rightarrow a + b = 2 A . .. (i) \\ ∵ G . M . = \sqrt{ab} \\ \Rightarrow G = \sqrt{ab} \\ \Rightarrow G^{2} = ab . .. (ii) \\ ∵ a - b = \sqrt{{(a + b)}^{2} - 4 ab} \\ = \sqrt{{(2 A)}^{2} - 4 G^{2}} \\ a - b = 2 \sqrt{A^{2} - G^{2}} . .. (iii) \\ Solving equation (i) and equation (iii), we get \\ a = A + \sqrt{A^{2} - G^{2}} and b = A - \sqrt{A^{2} - G^{2}} \\ Therefore, the two numbers are: \\ A \pm \sqrt{(A + G) (A - G)} . \end{array}$

Q.26 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^nd hour, 4^th hour and n^th hour?

Ans.

$\begin{array}{l} Originally present bacteria in a culture = 30 \\ Number of bacteria after one hour = 60 \\ Since, number of bacteria doubles every hour. \\ So, common ratio (r) in G.P. = 2 \\ First term (a) in G.P. = 30 \\ Number {of bacteria after 2}^{nd} hour \\ = {ar}^{2} \\ = 30 \times 2^{2} \\ = 120 \\ Number {of bacteria after 4}^{th} hour \\ = {ar}^{4} \\ = 30 \times 2^{4} \\ = 480 \end{array}$

$\begin{array}{l} Number {of bacteria after n}^{th} hour \\ = {ar}^{n} \\ = 30 \times 2^{n} \\ = 30 (2^{n}) \end{array}$

Q.27 What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Ans.

$\begin{array}{l} Principal amount deposited in the bank = ₹ 500 \\ Rate of interest in the bank = 10 % p . a . \\ Number of years for amount deposited = 10 years \\ Amount after one year in the bank = 500 (1 + \frac{10}{100}) \\ = 500 (\frac{11}{10}) \\ Amount after two years in the bank = 500 {(1 + \frac{10}{100})}^{2} \\ = 500 {(\frac{11}{10})}^{2} \\ Amount after three years in the bank = 500 {(1 + \frac{10}{100})}^{3} \\ = 500 {(\frac{11}{10})}^{3} \\ So, G.P. formed by using amount obtained from bank is \end{array}$

$\begin{array}{l} 500 (\frac{11}{10}), 500 {(\frac{11}{10})}^{2}, 500 {(\frac{11}{10})}^{3}, . .. \\ First term of G.P. = 500 (\frac{11}{10}) \\ Common ratio of G.P. = \frac{11}{10} \\ Number of years = 10 \\ Since, { T}_{n} = {ar}^{n - 1} \\ \Rightarrow T_{10} = 500 (\frac{11}{10}) \times {(\frac{11}{10})}^{10 - 1} \\ = 500 {(1 .1)}^{10} \\ Thus, ₹ 500 amounts to ₹ 500 {(1 .1)}^{10} in 10 years after its \\ deposit in a bank. \end{array}$

Q.28 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Ans.

$\begin{array}{l} Let roots of a quadratic equation be a and b. \\ A.M. of a and b = 8 \\ \Rightarrow \frac{a + b}{2} = 8 \\ \Rightarrow a + b = 16 . .. (i) \\ G . M . of a and b = 5 \\ \Rightarrow \sqrt{ab} = 5 \\ \Rightarrow ab = 25 . .. (ii) \\ Quadratic equation, whose roots are a and b, \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} (x - a) (x - b) = 0 \\ x^{2} - (a + b) x + ab = 0 \\ x^{2} - 16 x + 25 = 0 [From equation (i) and (ii)] \\ Thus, the required quadratic equation is x^{2} - 16 x + 25 = 0 . \end{array}$

Please register to view this section

FAQs (Frequently Asked Questions)

1. Have Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 been classified topic-wise in accordance with the NCERT book of Mathematics?

Yes, Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been written topic-by-topic in accordance with the NCERT Mathematics book. Extramarks always ensures that NCERT books remain at the centre of the process of learning. Keeping this in mind, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been made accessible by Extramarks.

2. How can students access Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

Students can get Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 from the Extramarks website and mobile application, and also access and download the PDF format. These solutions are simple to understand and beneficial to students as they are preparing for board examinations and other competitive examinations.

3. Are Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 inclusive of the updated syllabus of Mathematics for Class 11?

Yes, Extramarks keeps track of all updates and new topics added to the syllabus. It goes the extra mile in order to ensure that students face no difficulties. As a result, the Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 have been compiled while heeding the nuances of the latest syllabus of Mathematics for Class 11.

4. Is it helpful to download Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

Yes, without a doubt, downloading Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 are very beneficial. These solutions have been organised topic-wise and contain detailed calculations and applications of formulae to solve questions which have been given as a part of Exercise 9.3 of Chapter 9 of the NCERT textbook of Mathematics. These solutions could also be used in the form of revision notes during examinations. These are excellent reference materials which can be readily accessed from the Extramarks educational website.

5. Why are NCERT Solutions for Mathematics important?

With the help of real-world examples, NCERT Solutions for Chapter 9 of the Mathematics syllabus for Class 11, teach students how to understand the topic of Sequences in the most basic way possible. The content in NCERT books has been compiled through extensive research and painstaking efforts which makes them indispensable for learning. The NCERT Solutions would help students in tapping the infinite learning potential of the NCERT-prescribed academic curriculum for Mathematics.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

Access NCERT Solutions for Class 11 Maths Chapter 9 –Sequences and Series

Exercise 9.3

NCERT Solutions for Class 11 Maths Chapters

NCERT Solution Class 11 Maths of Chapter 9 Exercises

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3

Topics Covered in Exercise 9.3 of Chapter 9 Sequences and Series

Please register to view this section

FAQs (Frequently Asked Questions)

1. Have Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 been classified topic-wise in accordance with the NCERT book of Mathematics?

2. How can students access Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

3. Are Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 inclusive of the updated syllabus of Mathematics for Class 11?

4. Is it helpful to download Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

5. Why are NCERT Solutions for Mathematics important?

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.3) Exercise 9.3

Access NCERT Solutions for Class 11 Maths Chapter 9 –Sequences and Series

Exercise 9.3

NCERT Solutions for Class 11 Maths Chapters

NCERT Solution Class 11 Maths of Chapter 9 Exercises

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3

Topics Covered in Exercise 9.3 of Chapter 9 Sequences and Series

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. Have Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 been classified topic-wise in accordance with the NCERT book of Mathematics?

2. How can students access Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

3. Are Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 inclusive of the updated syllabus of Mathematics for Class 11?

4. Is it helpful to download Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3?

5. Why are NCERT Solutions for Mathematics important?

Share

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions