# 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.

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, a_{1} = −10, a_{2} = − 6, a_{3} = − 2 and a_{4} = 2,

Now, d_{1 }= a_{2 }– a_{1 }= − 6 − (−10) = − 6 + 10 = 4

d_{2} = a_{3} – a_{1} = − 2 − (− 6) = − 2 + 6 = 4

And d_{3} = a_{4} - a_{3} = 2 − (− 2) = 2 + 2 = 4

As d_{1} = d_{2} = d_{3 }= 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, a_{1} = –2 and d = –2

Therefore, a_{2} = a_{1} + d

⟹ a_{2} = –2 – 2 = – 4

Also a_{3} = a_{2 }+ d = – 4 + (–2) = –6

And a_{4} = a_{3} + 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 a_{1}= −3 + d = 4

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

Therefore, a_{21} = 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, a_{2} = 13 and a_{5} = 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

⟹ a_{7 }= 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 a_{18} - a_{13}?

(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:

a_{18 }=_{ }a + (18 − 1) d

And a_{13} = a + (13 − 1) d

Thus, a_{18} – a_{13} = [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**

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