NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 PDF

A sequence is an ordered list of numbers following a specific pattern, while a series is the sum of terms in a sequence. This exercise mainly deals with the nth term and sum of n terms in an arithmetic progression (AP).These concepts help in solving real-life problems involving patterns and progressions efficiently.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 Overview

NCERT Solutions for Class 11 Maths Chapter 8, Exercise 8.1 provide an in-depth understanding of Sequences and Series. This exercise introduces the concept of sequences, which are ordered lists of numbers following a specific rule, and series, which are the sums of sequence terms. The primary focus is on Arithmetic Progressions (APs), including finding the nth term and the sum of n terms. These concepts are crucial for understanding patterns, financial calculations, and other real-world applications.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1  PDF

Below, we have provided the NCERT Solutions for Class 11 Maths Chapter 8, Exercise 8.1 in PDF format. This exercise covers the fundamental concepts of Sequences and Series, focusing on Arithmetic Progressions (APs), including the nth term formula and the sum of n terms formula. These solutions are designed to help students understand and solve problems efficiently. Click the link below to download the NCERT Solutions PDF and enhance your learning with step-by-step explanations and solved examples.

NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1  PDF

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

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

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

1. an = n (n + 2) 

Solution:

Given,

nth term of a sequence an = n (n + 2) 

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a1 = 1(1 + 2) = 3

a2 = 2(2 + 2) = 8

a3 = 3(3 + 2) = 15

a4 = 4(4 + 2) = 24

a5 = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

2. an = n/n+1

Solution:

Given the nth term, an = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

3. an = 2n

Solution:

Given the nth term, an = 2n

On substituting n = 1, 2, 3, 4, 5, we get

a1 = 21 = 2

a2 = 22 = 4

a3 = 23 = 8

a4 = 24 = 16

a5 = 25 = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

4.  an = (2n – 3)/6

Solution:

Given the nth term, an = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

5. an = (-1)n-1 5n+1

Solution:

Given the nth term, an = (-1)n-1 5n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, –125, 625, –3125, and 15625.

6.

Solution:

On substituting n = 1, 2, 3, 4, 5, we get the first 5 terms.

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

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

7. an = 4n – 3; a17, a24

Solution:

Given,

The nth term of the sequence is an = 4n – 3

On substituting n = 17, we get

a17 = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a24 = 4(24) – 3 = 96 – 3 = 93

8. an = n2/2n ; a7

Solution:

Given,

The nth term of the sequence is an = n2/2n

Now, on substituting n = 7, we get

a7 = 72/27 = 49/ 128

9. an = (-1)n-1 n3; a9

Solution:

Given,

The nth term of the sequence is an = (-1)n-1 n3

On substituting n = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

10.

Solution:

On substituting n = 20, we get

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

11. a1 = 3, an = 3an-1 + 2 for all n > 1

Solution:

Given, an = 3an-1 + 2 and a1 = 3

Then,

a2 = 3a1 + 2 = 3(3) + 2 = 11

a3 = 3a2 + 2 = 3(11) + 2 = 35

a4 = 3a3 + 2 = 3(35) + 2 = 107

a5 = 3a4 + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

12. a1 = -1, an = an-1/n, n ≥ 2

Solution:

Given,

an = an-1/n and a1 = -1

Then,

a2 = a1/2 = -1/2

a3 = a2/3 = -1/6

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

13. a1 = a2 = 2, an = an-1 – 1, n > 2

Solution:

Given,

a1 = a2, an = an-1 – 1

Then,

a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

14. The Fibonacci sequence is defined by

1 = a1 = a2 and an = an – 1 + an – 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5 

Solution:

Given,

1 = a1 = a2

an = an – 1 + an – 2, n > 2

So,

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

Thus,

Benefits of Using NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 Sequences and Series

Concept Clarity – Provides a clear understanding of sequences and series, especially arithmetic progressions (APs).

Step-by-Step Solutions – Detailed explanations help students grasp problem-solving techniques effectively.

Exam-Oriented Approach – Covers important formulas like the nth term and sum of n terms to aid in exam preparation.

Error-Free and Accurate – Solutions are prepared by experts, ensuring correctness.

Time-Saving – Helps in quick revision and better time management during exams.

Boosts Problem-Solving Skills – Enhances logical thinking and mathematical reasoning.