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

Here 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 a_n = 4, then a is equal to

(a) 6

(b) 7

(c) 20

(d) 28

Solution. (d)

Explanation:

We know that in an AP,

     a_n = 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 a_n will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Solution. (b)

Explanation:

We know that in an AP,

      a_n = a + (n − 1) d

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

⟹ a_n = 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.

(a) –20

(b) 20

(c) –30

(d) 30

Solution. (b)

Explanation:

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, a_n = 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 a_n = 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) 9^th

(b) 10^th

(c) 11^th

(d) 12^th

Solution. (b)

Explanation:

Let nth term of the given AP be 210.

Here, first term, a = 21

Common difference, d = 42 − 21 = 21

And a_n = 210

Using a_n = 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 a_n = 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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