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

In this article you will get CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IB). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Created On: Jul 4, 2017 10:33 IST
Modified On: Jul 7, 2017 13:18 IST Here 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:

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:

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:

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

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NCERT Solutions for CBSE Class 10 Maths

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