Study MaterialsNCERT SolutionsNCERT Solutions for Class 11 MathematicsNCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1

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

The NCERT Solutions for the first exercise of Class 11 Maths Chapter 9 are available here and can also be downloaded in PDF format. Chapter 9, Sequences and Series, is part of the CBSE Syllabus for 2024-25. Exercise 9.1 of NCERT Solutions for Class 11 Maths covers the following topics:

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    • Introduction to Sequences and Series
    • Sequences
    • Series

    The NCERT textbook includes numerous questions for students to solve and practice. To achieve high marks in the Class 11 examination, it is essential to solve and practice the NCERT Solutions for Class 11 Maths.

    NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1

    Download the free PDF of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1, prepared by expert Mathematics teachers at Infinity Learn, following CBSE (NCERT) guidelines. Access our Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 solutions to help you revise the entire syllabus and score higher marks in your exams.

    Access NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

    1. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are: 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. Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

    1. an = 4n – 3; a17, a24

    Solution:

    Given,

    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

    2. an = n2/2n ; a7

    Solution:

    Given,

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

    Now, on substituting n = 7, we get

    a7 = 72/27 = 49/ 128

    3. 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) + ……

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