Jagran Josh Logo
  1. Home
  2. |  
  3. CBSE Board|  

NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IB)

Jul 4, 2017 10:33 IST

    Class 10 Maths NCERT Exemplar, Arithmetic Progressions NCERT Exemplar Problems, NCERT Exemplar Problems, Class 10 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 5, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-IB). This part of the chapter includes solutions of Question Number 10 to 18 from Exercise 5.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution.

    NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IA)

    NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

    Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:

    Exercise 5.1

    Multiple Choice Questions (Q. No. 9-18)

    Question. 10 What is the common difference of an AP in which a18 − a14 = 32?

    (a) 8

    (b) – 8

    (c) – 4

    (d) 4

    Solution. (a)

    Explanation:

    Given, a18 - a14 = 32

    Using an = a + (n – 1)d, we have:

                a18 =  a + (18 − 1) d

    And     a13 = a + (13 − 1) d

    Thus, a18 – a13 = [a + (18 − 1)d] − [a + (14 − 1) d] = 32

    ⟹       a + 17d – a – 13d = 32

    ⟹       4d = 32

    ⟹       d = 8

    Thus, the required common difference of given AP = 8.

    Question. 11 Two APs have the same common difference. The first term of one of these is − 1 and that of the other is − 8. The difference between their 4th terms is

    (a) - 1

    (b) – 8

    (c) 7

    (d) - 9

    Solution. (c)

    Explanation:

    Let the same common difference of two APs be d.

    Given, first term of first AP (a1) = − 1

    and the first term of second AP (a1’) = − 8

    We know that, the nth term of an AP is given as,

                an = a + (n − 1) d

    Therefore, 4th term of first AP is,

                a4 = a1 + (4 − 1) d = −1 + 3d

    And 4th term of second AP is,

                a4’ = a1’ + (4 − 1) d = − 8 + 3d

    Difference between 4th terms of both APs is:

                a4 − a4’ = (−1 + 3d) − (− 8 + 3d)

                            = − 1 + 3d + 8 − 3d = 7

    Hence, the required difference is 7.

    Question. 12 If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

    (a) 7

    (b) 11

    (c) 18

    (d) 0

    Solution. (d)

    Explanation:

    We know that, the nth term of an AP is given as,

                an = a + (n − 1) d

    Now, according to the question,

                                     7a7 = 11a11

    ⟹       7 [a + (7 − 1) d] = 11 [a + (11-1) d]                           

    ⟹                  7 (a +6d) = 11 (a + 10d)

    ⟹                  7a + 42d = 11a + 110d

    ⟹                   4a + 68d = 0

    ⟹                4 (a +17d) = 0

    ⟹                       a + 7d = 0                            [4 ≠ 0]

    Or              a + (18 - 1) d = 0  

    ⟹18th term of an AP, a18 = 0

    Question. 13 The 4th term from the end of an AP − 11, − 8, − 5,..., 49  is

    (a) 37

    (b) 40

    (c) 43

    (d) 58

    Solution. (b)

    Explanation:

    Taking the AP in reverse order: 49,…, −5, −8, −11.

    Here, al = 49   

    Common difference, d = − 8 − (− 11)

                                           = − 8 + 11= 3

    Now, we know that, the nth term of an AP is given as:

                an = al – (n − 1) d       

    Therefore, fourth term of AP is,

                a4 = 49 − (4 − 1) 3 = 49 − 9 = 40

    Question. 14 The famous mathematician associated with finding the sum of the first 100 natural numbers is

    (a) Pythagoras

    (b) Newton

    (c) Gauss

    (d) Euclid

    Sol. (c)

    Explanation:

    Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., 1 + 2 + 3 +...+ 100.

    Question. 15 If the first term of an AP is − 5 and the common difference is 2, then the sum of the first 6 terms is

    (a) 0

    (b) 5

    (c) 6

    (d) 15

    Solution. (a)

    Explanation:

    Arithmetic Progressions MCQs

    Question. 16 The sum of first 16 terms of the AP 10, 6, 2,… is

    (a) −320

    (b) 320

    (c) −352

    (d) −400

    Solution. (a)

    Explanation:

    Arithmetic Progressions NCERT Exemplar Problems

    Question. 17 In an AP, if a = 1, an = 20 and Sn = 399, then n is equal to

    (a) 19

    (b) 21

    (c) 38

    (d) 42

    Solution. (c)

    Explanation:

    Arithmetic Progressions Class 10

    Question. 18 The sum of first five multiples of 3 is

    (a) 45

    (b) 55

    (c) 65

    (d) 75

    Solution. (a)

    Explanation:

    The first five multiples of 3 are 3, 6, 9, 12 and 15.

    3, 6, 9, 12 and 15 form an AP with both its first term and common difference (6 – 3) being 3.

    i.e., a = 3 and d = 3

    Also, number of terms, n = 5

    Arithmetic Progressions NCERT MCQs

    You may also like to read:

    CBSE Class 10 Mathematics Syllabus 2017-2018

    CBSE Class 10 NCERT Textbooks & NCERT Solutions

    NCERT Solutions for CBSE Class 10 Maths

    NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters

    DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article.

    Latest Videos

    Register to get FREE updates

      All Fields Mandatory
    • (Ex:9123456789)
    • Please Select Your Interest
    • Please specify

    • ajax-loader
    • A verifcation code has been sent to
      your mobile number

      Please enter the verification code below

    Newsletter Signup
    Follow us on
    This website uses cookie or similar technologies, to enhance your browsing experience and provide personalised recommendations. By continuing to use our website, you agree to our Privacy Policy and Cookie Policy. OK
    X

    Register to view Complete PDF