NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

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To have a better understanding of the exercise questions, download Topper’s NCERT solutions for class 10 maths chapter 5. Also, with the Toppr learning app, you will get to attend the free conceptual videos and master classes live. Not only that, but you will also have access to all the free PDFs for solutions and study materials. So, download the Toppr learning app now!

NCERT class 10 maths arithmetic progression is a part of unit 2 in algebra. In this chapter, students will also discuss patterns in which they are succeeding to obtain by adding a fixed number for the preceding terms. Also, they will see about finding the nth and sum of n consecutive terms. So, this knowledge can be useful in solving some daily life problems. The chapter talks about dealing with the introduction to arithmetic progression with the help of some daily life problems. Also, there are simple and easy problems given at regular intervals to ease the concepts.

The examples, pictorial representation, lucid language, and highlighting the main terms make this chapter more interesting and interactive. Also, there is a summary of the concepts given at the end that will help understand the chapter at a glance. This part of unit 2, arithmetic progression, holds a total weight of 20 marks in the main exam.

You will get a 100% accurate NCERT solutions for maths class 10 chapter 5 solved by our maths subject experts. We will also provide step by step solution to the questions given in the maths textbook as per the NCERT guidelines. Download Toppr for Android and iOS or signup for free.

Arithmetic Progression Sub-topics

• Ex 5.1 – Introduction to Arithmetic Progressions
• Ex 5.2 – Arithmetic Progressions
• Ex 5.3 – nth Term of an AP
• Ex 5.4 – Sum of First n Terms of an AP
• Ex 5.5 – Summary

You can download the complete NCERT Solutions of Class 10 Maths Chapter 5 by clicking on the button below.

NCERT Solutions for Class 10 Maths Chapter 5

Q-1 Which term of the AP : 3, 8, 13, 18, . . . , is 78?

Solution:-

Given A.P. is 3, 8, 13, 18, …

For the above AP,

a = 3

d = a2 − a1 = 8 − 3 = 5

Let nth term of this A.P. be 78.

an = a + (n − 1)d

78 = 3 + (n − 1)5

75 = (n − 1)5

(n − 1) = 15

n = 16

Hence, 16th term of this A.P. is 78.

Q-2 Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Solution:-

Given that,

a11 = 38

a16 = 73

We know that,

an = a + (n − 1)d

a11 = a + (11 − 1)d

38 = a + 10d … (i)

Similarly,

a16 = a + (16 − 1)d

73 = a + 15d … (ii)

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

35 = 5d

d = 7

From equation (i),

38 = a + (10)(7)

38 − 70 = a

a = − 32

∴ a31 = a + (31 − 1)d

= − 32 + 30(7)

= − 32 + 210

= 178

Hence, 31st term is 178.

Q-3 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:-

Given, a3 = 12, a50 = 106

an = a + (n − 1)d

a3 = a + (3 − 1)d

12 = a + 2d … (i)

a50 = a + (50 − 1)d

106 = a + 49d … (ii)

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

94 = 47d

d = 2

From equation (i), we get

12 = a + 2(2)

a = 12 − 4 = 8

∴ a29 = a + (29 − 1)d

= 8 + (28)2

= 8 + 56

= 64

Therefore, 29th term is 64.

Q-4 If the 3rd and the 9th terms of an AP are 4 and −8 respectively, then which term of this AP is zero.

Solution:-

Given that, a3 = 4, a9 = − 8

We know that,

an = a + (n − 1)d

a3 = a + (3 − 1)d

4 = a + 2d … (i)

And

a9 = a + (9 − 1)d

−8 = a + 8d … (ii)

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

−12 = 6d

d = − 2

From equation (i), we get,

4 = a + 2( − 2)

⇒ a = 8

Let nth term of this A.P. be zero.

These are some of the examples of solutions given in the free PDF prepared by our subject experts. For more detailed solutions and questions are given in the textbook download the PDF.

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