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

In this article you will get CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IA). 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.

Jul 3, 2017 17:21 IST
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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-IA). This part of the chapter includes solutions of Question Number 1 to 9 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 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. 1-9)

Question. 1 In an AP, if d = − 4, n = 7 and an = 4, then a is equal to

(a) 6

(b) 7

(c) 20

(d) 28

Solution. (d)

Explanation:

We know that in an AP,

     an = a + (n − 1) d

⟹ 4 = a + (7 − 1) (− 4)                       [by given conditions]

⟹ 4 = a + 6 (− 4)

⟹ a = 4 + 24 = 28

Question. 2 In an AP, if a = 3.5, d = 0 and n = 101, then an will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Solution. (b)

Explanation:

We know that in an AP,

      an = a + (n − 1) d

⟹ an = 3.5 + (101 − 1) × 0     [by given conditions]

⟹ an = 3.5

Question. 3 The list of numbers − 10, − 6, − 2, 2, ... is

(a) an AP with d = −16

(b) an AP with d = 4

(c) an AP with d = − 4

(d) not an AP

Solution. (b)

Explanation:

Given list of numbers is: −10, − 6, − 2, 2,....

Here, a1 = −10, a2 = − 6, a3 = − 2 and a4 = 2,

Now, d1 = a2 – a1 = − 6 − (−10) = − 6 + 10 = 4

d2 = a3 – a1 = − 2 − (− 6) = − 2 + 6 = 4

And d3 =  a4 - a3 = 2 − (− 2) = 2 + 2 = 4

As d1 = d2 = d3 = 4

So, the given list forms an AP with common difference, d = 4.

Arithmetic Progressions NCERT Exemplar Problems

(a) –20

(b) 20

(c) –30

(d) 30

Solution. (b)

Explanation:

Arithmetic Progressions NCERT Exemplar Problems, NCERT Exemplar Problems

Question. 5 The first four terms of an AP whose first term is − 2 and the common difference is − 2 are

(a) − 2, 0, 2, 4

(b) − 2, 4, − 8, 16

(c) − 2, − 4, − 6, − 8

(d) − 2, − 4, − 8, − 16

Solution. (c)

Explanation:

Given, a1 = –2 and  d = –2

Therefore, a2 = a1 + d

⟹       a2 = –2 – 2 = – 4

Also     a3 = a2 + d = – 4 + (–2) = –6

And     a4 = a3 + d = – 6 + (–2) = –8

So, the first four terms are –2, –4, –6, –8.

Question. 6 The 21st term of an AP whose first two terms are − 3 and 4, is

(a) 17

(b) 137

(c) 143

(d) −143

Solution. (b)

Explanation:

For an AP, an = a + (n – 1)d  …..(i)

Given, first two terms of an AP are − 3 and a + d = 4.

 a = − 3 and a1= −3 + d = 4

⟹ Common difference, d = a1 – a = 4 – (−3) = 4+3 = 7

Therefore, a21 = a + (21 − 1) d            [Using (i)]

                       = − 3 + (20) 7

                       = − 3 + 140 = 137

Question. 7 If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term?

(a) 30

(b) 33

(c) 37

(d) 38

Solution. (b)

Explanation:

Given, a2 = 13 and a5 = 25

Using an = a + (n – 1)d, we have:

      a + (2 − 1) d = 13 and      a + (5 − 1) d = 25

⟹  a + d = 13 ….(i)  and a + 4 d = 25   …..(ii)

On subtracting equation (i) from equation (ii), we get

     3 d = 25 – 13 = 12

⟹  d = 4

Putting value of d in equation (i), we get:

       a = 13 − 4 = 9

⟹  a7 = a + (7 − 1) d = 9 + 6 × 4 = 33

Question. 8 Which term of an AP : 21, 42, 63, 84, ... is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Solution. (b)

Explanation:

Let nth term of the given AP be 210.

Here, first term, a = 21

Common difference, d = 42 − 21 = 21

And an = 210

Using an = a + (n – 1)d, we have:

       210 = 21 + (n - 1) 21

⟹   210 = 21+ 21n – 21

⟹   210 = 21 n

⟹      n = 10

Hence, the 10th term of an AP is 210.

Question. 9 If the common difference of an AP is 5, then what is a18 - a13?

(a) 5

(b) 20

(c) 25

(d) 30

Solution. (c)

Explanation:

Given, common difference of an AP, d = 5

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 + (13 -1) d]                   

                        = a + 17 × 5 – a – 12 × 5

                        = 85 – 60 = 25

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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

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