NCERT Solutions for Exercise 9.1 Class 11 Maths Chapter 9 - Sequences and Series

# NCERT Solutions for Exercise 9.1 Class 11 Maths Chapter 9 - Sequences and Series

Edited By Vishal kumar | Updated on Nov 08, 2023 07:40 AM IST

## NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.1- In simple words 'sequence' means the collections of objects in order such that these objects are identified as the first member, second member, third member, and so on. In the NCERT solutions for Class 11 Maths chapter 9 exercise 9.1, you will get questions related to sequences. Class 11 Maths chapter 9 exercise 9.1 solutions consist of questions related to series also. The sum of a sequence of terms is called the series and the sequence following the specific pattern is called progression.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

In NCERT syllabus of previous classes, you have already learned about arithmetic progression. As you are familiar with this chapter, it won't take much effort to understand this chapter. You are advised to solve NCERT book Class 11 Maths chapter 9 exercise 9.1 problems to get conceptual clarity. The Class 11 Maths chapter 9 exercise 9.1 is the basic exercise of this chapter where you will learn about the basic definitions and few problems of sequence and series. You can go through the Class 11 Maths chapter 9 exercise 9.1 solutions, so you can solve the NCERT problems without any difficulty. Check NCERT Solutions from Classes 6 to 12 for Science and Maths at one place.

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**As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 8.

## Question:1 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

$a _n = n ( n +2)$

Given : $a _n = n ( n +2)$

$a _1 = 1 ( 1 +2)=3$

$a _2 = 2 ( 2 +2)=8$

$a _3 = 3 ( 3 +2)=15$

$a _4 = 4 ( 4 +2)=24$

$a _5 = 5 ( 5 +2)=35$

Therefore, the required number of terms =3, 8, 15, 24, 35

$a _n = \frac{n }{n+1}$

Given : $a _n = \frac{n }{n+1}$

$a _1 = \frac{1}{1+1}=\frac{1}{2}$

$a _2 = \frac{2}{2+1}=\frac{2}{3}$

$a _3 = \frac{3}{3+1}=\frac{3}{4}$

$a _4 = \frac{4}{4+1}=\frac{4}{5}$

$a _5 = \frac{5}{5+1}=\frac{5}{6}$

Therefore, the required number of terms $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$

$a _ n = 2 ^n$

Given : $a _ n = 2 ^n$

$a _ 1 = 2 ^1=2$

$a _ 2 = 2 ^2=4$

$a _ 3 = 2 ^3=8$

$a _ 4 = 2 ^4=16$

$a _ 5 = 2 ^5=32$

Therefore, required number of terms $=2,4,8,16,32.$

$a _n = \frac{2n-3 }{6}$

Given : $a _n = \frac{2n-3 }{6}$

$a _1 = \frac{2\times 1-3 }{6}=\frac{-1}{6}$

$a _2 = \frac{2\times 2-3 }{6}=\frac{1}{6}$

$a _3 = \frac{2\times 3-3 }{6}=\frac{3}{6}=\frac{1}{2}$

$a _4 = \frac{2\times 4-3 }{6}=\frac{5}{6}$

$a _5 = \frac{2\times 5-3 }{6}=\frac{7}{6}$

Therefore, the required number of terms $=\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

$a _ n = ( -1) ^{n-1} 5 ^{n+1}$

Given : $a _ n = ( -1) ^{n-1} 5 ^{n+1}$

$a _ 1 = ( -1) ^{1-1} 5 ^{1+1}=(-1)^{0}.5^2=25$

$a _ 2 = ( -1) ^{2-1} 5 ^{2+1}=(-1)^{1}.5^3=-125$

$a _ 3 = ( -1) ^{3-1} 5 ^{3+1}=(-1)^{2}.5^4= 625$

$a _ 4 = ( -1) ^{4-1} 5 ^{4+1}=(-1)^{3}.5^5= -3125$

$a _ 5 = ( -1) ^{5-1} 5 ^{5+1}=(-1)^{4}.5^6= 15625$

Therefore, the required number of terms $=25,-125,625,-3125,15625$

$a _n = n \frac{n^2 + 5}{4}$

Given : $a _n = n \frac{n^2 + 5}{4}$

$a _1 = 1. \frac{1^2 + 5}{4}=\frac{6}{4}=\frac{3}{2}$

$a _2 = 2. \frac{2^2 + 5}{4}=\frac{18}{4}=\frac{9}{2}$

$a _3 = 3. \frac{3^2 + 5}{4}=\frac{42}{4}=\frac{21}{2}$

$a _4 = 4. \frac{4^2 + 5}{4}=\frac{84}{4}=21$

$a _5 = 5. \frac{5^2 + 5}{4}=\frac{150}{4}=\frac{75}{2}$

Therefore, the required number of terms $=\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$

$a _ n = 4 n - 3 ; a _{17} , a _{24}$

$a _ n = 4 n - 3$

Put $n=17,$

$a {1_7 }= 4 (17) - 3=68-3=65$

Put n=24,

$a _ {24 }= 4 (24) - 3=96-3=93$

Hence, we have $a_{17}=65\, \, and\, \, a _ {24 }=93$

$a _n = \frac{n^2 }{2^n } ; a_7$

Given : $a _n = \frac{n^2 }{2^n }$

Put n=7,

$a _7 = \frac{7^2 }{2^7 } =\frac{49}{128}$

Heence, we have $a _7 =\frac{49}{128}$

$a _ n = ( -1) ^{n-1} n ^ 3 , a _9$

Given : $a _ n = ( -1) ^{n-1} n ^ 3$

Put n =9,

$a _ 9 = ( -1) ^{9-1} 9 ^ 3= (1).(729)=729$

The value of $a _ 9 =729$

$a _n = \frac{n ( n-2)}{ n+3 }; a _{20}$

Given : $a _n = \frac{n ( n-2)}{ n+3 }$

Put n=20,

$a _2_0 = \frac{20 ( 20-2)}{ 20+3 }=\frac{360}{23}$

Hence, value of $a _2_0=\frac{360}{23}$

$a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$

Given : $a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$

$a_2 = 3a_{2 - 1} + 2=3a_1+2=3(3)+2=11$

$a_3 = 3a_{3 - 1} + 2=3a_2+2=3(11)+2=35$

$a_4 = 3a_{4 - 1} + 2=3a_3+2=3(35)+2=107$

$a_5 = 3a_{5 - 1} + 2=3a_4+2=3(107)+2=323$

Hence, five terms of series are $3,11,35,107,323$

Series $=3+11+35+107+323+...............$

$a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$

Given : $a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$

$a _ 2 = \frac{a_{2-1}}{2} =\frac{a_1}{2}=\frac{-1}{2}$

$a _ 3 = \frac{a_{3-1}}{3} =\frac{a_2}{3}=\frac{-1}{6}$

$a _ 4 = \frac{a_{4-1}}{4} =\frac{a_3}{4}=\frac{-1}{24}$

$a _ 5 = \frac{a_{5-1}}{5} =\frac{a_4}{5}=\frac{-1}{120}$

Hence, five terms of series are $-1,\frac{-1}{2},\frac{-1}{-6},\frac{-1}{24},\frac {-1}{120}$

Series

$=-1+\frac{-1}{2}+\frac{-1}{-6}+\frac{-1}{24}+\frac {-1}{120}.........................$

Given : $a_1 = a_2 = 2, a_n = a_{n - 1}-1, n > 2$

$a_3 = a_{3 - 1}-1=a_2-1=2-1=1$

$a_4 = a_{4 - 1}-1=a_3-1=1-1=0$

$a_5 = a_{5 - 1}-1=a_4-1=0-1=-1$

Hence, five terms of series are $2,2,1,0,-1$

Series $=2+2+1+0+(-1)+..................$

Find$\frac{a _{n+1}}{a_n}$, for n = 1, 2, 3, 4, 5

Given : The Fibonacci sequence is defined by $1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2$

$a _3 = a _{3-1} + a _{3-2} =a_2+a_1=1+1=2$

$a _4 = a _{4-1} + a _{4-2} =a_3+a_2=2+1=3$

$a _5 = a _{5-1} + a _{5-2} =a_4+a_3=3+2=5$

$a _6 = a _{6-1} + a _{6-2} =a_5+a_4=5+3=8$

$For \,\,n=1,\frac{a _{n+1}}{a_n}=\frac {a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$

$For \,\, n=2,\frac{a _{n+1}}{a_n}=\frac {a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$

$For \,\, n=3,\frac{a _{n+1}}{a_n}=\frac {a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$

$For \,\, n=4,\frac{a _{n+1}}{a_n}=\frac {a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$

$For \,\, n=5,\frac{a _{n+1}}{a_n}=\frac {a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$

## More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1:-

Class 11 Maths chapter 9 exercise 9.1 consists of questions related to finding the nth term of the sequence, series of the given sequence, Fibonacci series, etc. There are some solved examples and few definitions related to sequence, series, and progression that are given before the Class 11 Maths chapter 9 exercise 9.1. You must go through these solved examples and definitions before solving the exercise problems. It will help you to understand the concept easily.

Also Read| Sequences And Series Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1:-

• You are advised to solve all the NCERT exercise problems by yourself. Exercise 9.1 Class 11 Maths solutions are here to help you when you are stuck while solving these probles.
• Class 11 Maths chapter 9 exercise 9.1 solutions are designed in a very simple language that is very easy to understand.
• You can use Class 11 Maths chapter 9 exercise 9.1 solutions as a reference while solving the NCERT problems.
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## Key Features of NCERT 11th Class Maths Exercise 9.1 Answers

1. Comprehensive Coverage: The ex 9.1 class 11 answers cover all the exercises and problems in exercise 9.1 of the NCERT 11th Class Mathematics textbook.

2. Step-by-Step Solutions: The class 11 maths ex 9.1 solutions are presented in a step-by-step format, making it easier for students to understand and follow the problem-solving process.

3. Clarity and Accuracy: The class 11 ex 9.1 answers are written with clarity and accuracy, ensuring that students can grasp the mathematical concepts and methods required to solve problems.

4. Proper Mathematical Notation: These 11th class maths exercise 9.1 answers use appropriate mathematical notations and terminology, helping students become familiar with the language of mathematics.

5. Conceptual Understanding: The ex 9.1 class 11 solutions focus on promoting a deep understanding of mathematical concepts rather than rote memorization, encouraging critical thinking and problem-solving skills.

6. Free Access: Typically, these exercise 9.1 class 11 maths answers are available free of charge, allowing students to access them without any cost, making it a valuable resource for self-study.

7. Supplementary Learning: These answers can be used as a supplementary learning resource to reinforce classroom learning and aid in exam preparation.

8. Homework and Practice: Students can use these answers to check their work, practice problem-solving, and improve their overall performance in mathematics.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Find the 2nd term of sequence whose nth term is An = n (n + 2) .

An  = n (n + 2)

A2 = 2(2+2)

A2 = 8

2. Find the 3rd term of sequence whose nth term is an =2^n .

An = 2^n

A3 = 2^3 = 8

3. What is an finite sequence ?

The finite sequence is a sequence that contains a finite number of terms.

4. What is an infinite sequence ?

The infinite sequence is a sequence in which the number of terms is not finite.

5. What is the weightage of sequence and series in the JEE Main exam ?

The weightage of sequence and series in JEE Main Maths is 6.6%. Generally, two questions from this chapter is asked in the JEE Main exam.

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