NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.1) Exercise 9.1

Mathematics is one of the most important subjects for Class 11 students. A basic understanding of Mathematics at primary school is critical for developing a student’s reasoning and logical thinking skills. Furthermore, Mathematics is used in almost every field in some way. As a result, understanding Mathematics is advantageous in many careers. There is a wide range of career alternatives open after studying Mathematics. Mathematics is taught in schools since the very beginning of elementary school and is one of the most important subjects in a curriculum. As a result, understanding the fundamentals of Mathematics is critical to a student’s developmental process.

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NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series (Ex 9.1) Exercise 9.1

Mathematical Progression and Geometric Progression are topics students will study in Class 11 Mathematics Chapter 9. Chapter 9 is challenging because of its formulas and tricks to solve the questions. Moreover, there are many exercises in NCERT of Chapter 9 for students to practice.

Class 11 Maths Chapter 9 Exercise 9.1 is based on the chapter’s introduction. Students need to learn and understand some formulas to solve this exercise. It is the first exercise of Chapter 9, so students might get confused or stuck in some questions. Students can take help from NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 available on the Extramarks’ website and mobile application.

In the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1, students will find all the solutions that will help them solve and understand the concepts of this exercise. Students should do a lot of practice in this chapter to score the best marks in exams. Moreover, to complete Exercise 9.1, students can use the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1. Extramarks offers live doubt-solving sessions on their website and mobile app, where students can ask questions about the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1.

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Access NCERT Solutions For Class 11 Maths Chapter 9 – Sequences And Series

Class 11 is the foundation of a student’s career journey, and it is important to clear the basis of Mathematics. Class 12 CBSE board is mostly the second level of Class 11 in Mathematics. Moreover, many students start preparing for the entrance exams along with Class 11. The entrance examination syllabus mostly consists of Class 11 and Class 12.

The NCERT Solutions for Class 11 Chapter 9 is about Sequences and Series. This chapter is new in Class 11 and slightly different from Class 10. Exercise 9.1 is the first exercise of chapter 9, and it is very important to understand the chapter to solve the exercises. Students can get help from the Extramarks. Extramarks is one of the best learning platforms that provides students with NCERT Solutions. Students can easily take the guidance of the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1, and start solving the exercises.

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NCERT Solutions For Class 11 Maths Chapter 9 Sequences And Series Exercise 9.1

In the chapter ‘Sequences and Series,’ students will discover Arithmetic Progression, Geometric Progression, the General Term of a G.P, sum to n terms of a G.P, and other key aspects. It is a required component of Class 11 Mathematics. The exercise in Chapter 9 is fascinating. Furthermore, students who are experiencing difficulties solving 9.1 in Class 11 can prefer to take help from the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1.

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Before preparing for the mathematics exams, taking care of a few points is important. Class 11 students must regularly practice the chapters from the NCERT textbooks. All chapter exercises must be completed regularly. It is critical to use online resources to learn the concepts effectively.

Q.1 Write the first five terms of the sequence whose nth terms is:
an = n(n+2)

Ans.

We havean= n(n+2)Substituting n=1,2,3,4,5 respectively, we geta1= 1(1+2)      =3a2= 2(2+2)      =8a3= 3(3+2)      =15a4= 4(4+2)      =24a5= 5(5+2)      =35Therefore, the first five terms are: 3, 8, 15, 24, 35.

Q.2

Write the first five terms of the sequence whose nth terms is:an=nn+1.

Ans.

We havean=nn+1Substituting n=1,2,3,4,5 respectively, we geta1=11+1      =12a2=22+1      =23a3= 33+1      =34a4=44+1      =45a5=55+1      =56Therefore, the first five terms are: 12, 23,34,45,56.

Q.3 Write the first five terms of the sequence whose nth terms is:
an = 2n

Ans.

We have, an= 2n Substituting n=1,2,3,4,5 respectively, we geta1= 21      =2a2= 22      =4a3= 23      =8a4=24      =16a5= 25      =32Therefore, the first five terms are: 2, 4, 8, 16, 32.

Q.4

Write the first five terms of the sequence whose nth terms is:an=2n−36.

Ans.

We havean=2n−36Substituting n=1,2,3,4,5 respectively, we geta1=2(1)−36      =−16a2=2(2)−36      =16

a3=2(3)−36      =36      =12a4=2(4)−36      =56a5=2(5)−36      =76Therefore, the first five terms are: −16,16,12,56,76.

Q.5 Write the first five terms of the sequence whose nth terms is an = (– 1)n–1 5n+1.

Ans.

We have an=(−1)n−15n+1Substituting n=1, 2, 3, 4, 5 respectively, we geta1=(–1)1−151+1      =25a2=(–1)2−152+1      =−125a3=(−1)3−153+1      =625

a4=(−1)4−154+1      =−3125a5=(−1)5−155+1      =15625Therefore, the first five terms are:25,−125,625,−3125,15625.

Q.6

Write the first five terms of the sequence whose nth terms isan=nn2+54

Ans.

We have an=nn2+54Substituting n = 1, 2, 3, 4, 5 respectively, we geta1=112+54      =64=32a2=222+54      =92a3=332+54      =424=212a4=442+54      =21a5=552+54      =1504      =752Therefore, the first five terms are: 32,  92,  212,  21,  752.

Q.7

Find the indicated terms in each of the sequences given below whose nth terms are:1. an= 4n–3; a17, a242. an=n22n;  a73. an= (–1)n–1n3; a94. an=n(n−2)n+3; a20

Ans.
1.

Given:an=4n–3Substituting n=17 and 24 respectively, we geta17=4(17)−3       =68−3       =65a24=4(24)−3       =96−3       =93

2.

Given:     an=n22nSubstituting n=7, we geta7=7227       =49128Thus, the value of a7 is 49128.

3.

Given:     an=(−1)n–1n3Substituting n=9, we get    a9=(−1)9–1(9)3 =(−1)8(729) =729Thus, the value of a9 is 729.

4. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A1E1@ Given:     an=n(n−2)n+3Substituting n=20, we get  a20=20(20−2)20+3

=20(18)23 =36023Thus, the value of a20 is 36023.

Q.8

Write the first five terms of each of the sequences given below andobtain the corresponding series:1. a1=3, an=3an-1+2, for all n>12. a1=–1, an=an-1n, n≥23. a1=a2=2, an=an–1–1, n > 2

Ans.

1.

Given:an=3an−1+2 and a1=3Substituting n=2, we geta2=3a2−1+2      =3a1+2      =3(3)+2[Putting a1=3]      =11Substituting n=3, we geta3=3a3−1+2      =3a2+2      =3(11)+2[Putting a2=11]      =35Substituting n=4, we geta4=3a4−1+2      =3a3+2      =3(35)+2[Putting a3=35]      =107

Substituting n=5, we geta5=3a5−1+2      =3a4+2      =3(107)+2[Putting a4=107]      =323Thus,​ the first five terms are 3, 11, 35, 107 and 323.Corresponding series is:  3+11+35+107+323+...

2.

Given:a1=−1,an=an–1n,n≥2Substituting n=2, we geta2=a2–12      =a12      =−12[Putting a1=−1]Substituting n=3, we geta3=a3–13      =a23      =(−12)3[Putting a2=−12]       

=−16Substituting n=4, we geta4=a4–14      =a34       

=(−16)4[Putting a3=−16]      =−124Substituting n=5, we geta5=a5–15      =a45      =(−124)5[Putting a4=−124]      =−1120Thus,​ the first five terms are −1, −12, −16, −124 and −1120.The corresponding series is:(−1)+(−12)+(−16)+(−124)+(−1120)+...

3.

Given:     an=an–1−1,  a1=a2=2, n>2Substituting n=3, we geta3=a3–1−1      =a2−1      =2−1[Putting a2=2]      =1

Substituting n=4, we geta4=a4–1−1      =a3−1      =1−1[Putting a3=1]      =0Substituting n=5, we geta5=a4–1−1      =a3−1      =0−1[Putting a4=0]      =−1Thus, the first five terms of series are:2, 2, 1, 0,−1The series is:  2+ 2+ 1+ 0+(−1)+...

Q.9

The Fibonacci sequence is defined by 1=a1=a2 and an=an−1+an−2,  n>2.Find an+1an, for n=1,2,3,4,5.

Ans.

The Fibonacci sequence is defined by1 = a1 = a2 and an= an−1+an−2, n>2.Substituting n = 3, we get                a3=a2+a1                     =1+1                     =2For n = 1, we get  a1+1a1=a2a1        =11     a2a1=1and a3a2=21=2     an+1an=an+an−1an-1+an−2Substituting n=3, we geta3+1a3=a3+a3−1a3−1+a3−2   a4a3=a3+a2a2+a1       =a2+a1+a2a2+a1    a3=a2+a1       =1+1+11+1      =32Substituting n=4, we get  a4=a3+a2      = 2+1      =3a4+1a4=a4+a3a3+a2  a5a4=3+22+1      =53Substituting n=5, we get                a5=a4+a3                    =3+2                    =5              a5+1a5=a5+a4a4+a3                a6a5=5+33+2                    =85Thus, the required terms are : 1,2,32,53,85.

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