NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 - Sequences and Series

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 - Sequences and Series

Komal MiglaniUpdated on 06 May 2025, 02:37 PM IST

The petals of a flower, the number of steps you climb each day, or even the pocket money you save weekly; they often follow a pattern called sequence. When the terms of a sequence are added together, they form a series. Here in this exercise, we will learn how the numbers are arranged in a certain order and how we calculate their sum.

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  1. Class 11 Maths Chapter 8 Exercise 8.1 Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 8: Exercise 8.1
  3. Topics covered in Chapter 8 Sequences and Series Exercise 8.
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions
NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 - Sequences and Series
NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 - Sequences and Series

The Solutions of NCERT will provide stepwise calculations with detailed explanations. These NCERT solutions will make tough-looking sums simple and will help you improve your accuracy. Whether you are figuring out the nth term or the logic behind a number pattern, these solutions will help you build a solid grip on the topic.


Class 11 Maths Chapter 8 Exercise 8.1 Solutions - Download PDF


Download PDF

NCERT Solutions Class 11 Maths Chapter 8: Exercise 8.1

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

$a _n = n ( n +2)$

Answer:

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

Question 2: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:

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

Answer:

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 $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$

Question 3: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

$a _ n = 2 ^n$

Answer:

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 $=2,4,8,16,32.$

Question 4: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

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

Answer:

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 $=\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

Question 5: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

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

Answer:

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 $=25,-125,625,-3125,15625$

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

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

Answer:

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 $=\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$

Question 7: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

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

Answer:

$a _ n = 4 n - 3$

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

Question 8: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are

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

Answer:

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

Put n=7,

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

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

Question 9: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

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

Answer:

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

Put n =9,

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

The value of $a _ 9 =729$

Question 10: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

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

Answer:

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

Put n=20,

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

Hence, value of $a_{20}=\frac{360}{23}$

Question 11: Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

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

Answer:

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

Question 12: Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

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

Answer:

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

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

Question 13: Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: $a_1 = a_2 = 2, a_n = a_{n - 1}-1, n > 2$

Answer:

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

Question 14: The Fibonacci sequence is defined by $1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2$

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

Answer:

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

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

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

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

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

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

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

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

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

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

Also read

Topics covered in Chapter 8 Sequences and Series Exercise 8.

1. Sequence - A sequence is an ordered list of numbers that follow a specific pattern or rule. We often write it using function notation like a(n) or an to represent the n-th term of the sequence.
For example: 2, 4, 6, 8, 10 is a sequence where each number increases by 2.

2. Finite sequence- A finite sequence is a sequence that has a limited number of terms.
For example: 3, 6, 9, 12 is a finite sequence with 4 terms.

3. Infinite sequence -An infinite sequence has no end.
For example: 1, 2, 3, 4, 5, 6, ... is an infinite sequence because it continues without stopping.

4. Series- A series is what you get when you add the terms of a sequence together.
If a sequence is in the form- $a_1, a_2, a_3, \ldots, a_n$, then the series corresponding to it is given by,

$$
a_1+a_2+a_3+\ldots+a_n+\ldots
$$

Also read

NCERT Solutions of Class 11 Subject Wise

Students can also check out the NCERT solutions for other subjects as well for effective learning.

Subject-Wise NCERT Exemplar Solutions

Use the links below to solve the NCERT exemplar solutions available for all the subjects. This will help you in your exam preparations.

Frequently Asked Questions (FAQs)

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

An  = n (n + 2) 

A2 = 2(2+2)

A2 = 8

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

An = 2^n

A3 = 2^3 = 8

Q: What is an finite sequence ?
A:

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

Q: What is an infinite sequence ?
A:

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

Q: What is the weightage of sequence and series in the JEE Main exam ?
A:


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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