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NCERT Solutions for Class 10 Maths Chapter 5 : Arithmetic Progressions

Arithmetic Progression class 10 Maths NCERT Solutions Chapter 5 is concerned with the understanding of different types of questions on arithmetic progressions. Students preparing for their Class 10 exams will be able to clear all their concepts on arithmetic progressions at the root level. Expert faculty of Toppr produced these solutions to assist students with their first term exam preparations. It covers all major concepts in detail, allowing students to understand the ideas better.

NCERT Solutions Arithmetic Progressions, the fifth chapter of the section, concentrates on the essential concepts such as the introduction to terminology like 'term,' common differences between terms, and the general shape of an arithmetic progression or an AP The 'n'-th term and the sum of 'n' successive terms of an arithmetic progression along with alternate solutions are also demonstrated using well-illustrated examples. All of these solutions are designed with the new CBSE pattern in mind so that students have a complete understanding of their tests.

Arithmetic Progressions, Chapter 5 Questions and Answers are very useful for getting good grades in tests and properly preparing you with all of the important concepts. These NCERT Solutions are valuable tools that can assist you not only in covering the full syllabus but also in providing an in-depth analysis of the subjects. The Class 10 Maths NCERT Solutions Chapter 5 are available in pdf format below, and some of them are also included in the exercises.

Table of Content

Introduction

Access NCERT Solutions for Class 10 Maths Chapter 5 : Arithmetic Progressions

Exercise 5.1

Question 1

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs.

15

for the first km and Rs.

8

for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes

\frac{1}{4}

of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs.

150

for the first metre and rises by Rs.

50

for each subsequent metre.
(iv) The amount of money in the account every year, when Rs.

10000

is deposited at compound interest at

8

% per annum.

Medium

Solution

Verified by Toppr

(i) Fare for first km

=

Rs.

15

Fare for second km $=$ Rs. $15 + 8 =$ Rs $23$

Fare for third km $=$ Rs. $23 + 8 = 31$

Here, each subsequent term is obtained by adding a fixed number $(8)$ to the previous term.

Hence, it is in A.P.

(ii) Let us assume, initial quantity of air $= 1$ .....1)

Therefore, quantity removed in first step $= \frac{1}{4}$

Remaining quantity after first step

$1 - \frac{1}{4} = \frac{3}{4}$ ....2)

Quantity removed in second step

=

\frac{3}{4} \times \frac{1}{4} = \frac{3}{1 6}

Remaining quantity after second step

=

\frac{3}{4} - \frac{3}{1 6} = \frac{9}{1 6}

....3)

Here, each subsequent term is not obtained by adding a fixed number to the previous term.

Hence, it is not an AP.

(iii) Cost of digging of $1^{s t}$ meter $= 150$

Cost of digging of $2^{n d}$ meter $= 150 + 50 = 200$

Cost of digging of $3^{r d}$ meter $= 200 + 50 = 250$

Here, each subsequent term is obtained by adding a fixed number $(50)$ to the previous term.

Hence, it is an AP.

(iv) Amount in the beginning $=$ Rs. $10000$

Interest at the end of $1^{s t}$ year @ $8 %$ = $10000 \times 8 % = 800$

Thus, amount at the end of $1^{s t}$ year $= 10000 + 800 = 10800$

Interest at the end of $2^{n d}$ year @ $8 %$ = $10800 \times 8 % = 864$

Thus, amount at the end of $2^{n d}$ year $= 10800 + 864 = 11664$

Since, each subsequent term is not obtained by adding a fixed number to the previous term; hence, it is not an AP.

Question 2

Write first four terms of the AP, when the first term

a

and the common difference

d

are given as follows:
(i)

a = 10, d = 10

(ii)

a = - 2, d = 0

(iii)

a = 4, d = - 3

(iv)

a = - 1, d =

\frac{1}{2}

(v)

a = - 1.25, d = - 0.25

Medium

Solution

Verified by Toppr

An arithmetic progression is given by

a, (a + d), (a + 2 d), (a + 3 d)

, .
where

a =

the first term,

d =

the common difference

(i)

10, 10 + 10, 10 + 2 (10)

and

10 + 3 (10) = 10, 20, 30

and

40

(ii)

- 2, - 2 + (0), - 2 + 2 (0)

and

- 2 + 3 (0) = - 2, - 2, - 2

and

- 2

(this is not an A.P)

(iii)

4, 4 + (- 3), 4 + 2 (- 3)

and

4 + 3 (- 3) = 4, 4 - 3, 4 - 6

and

4 - 9 = 4, 1, - 2

and

- 5

(iv)

- 1, - 1 + (\frac{1}{2}), - 1 + 2 (\frac{1}{2})

and

- 1 + 3 (\frac{1}{2}) = - 1, - \frac{1}{2}, 0

and

\frac{1}{2}

(v)

- 1.25, - 1.25 + (- 0.25), - 1.25 + 2 (- 0.25)

and

- 1.25 + 3 (- 0.25) = - 1.25, - 1.50, - 1.75

and

- 2.0

Question 3

For the following APs, write the first term and the common difference:
(i)

3, 1, - 1, - 3, . . . .

(ii)

- 5, - 1, 3, 7, . . . . .

(iii)

\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{1 3}{3}, . . . .

(iv)

0.6, 1.7, 2.8, 3.9, . . .

Medium

Solution

Verified by Toppr

(i)

3, 1, - 1, - 3 \dots

Here, first term,

a = 3

Common difference,

d =

Second term

-

First term

d = 1 - 3 = - 2

(ii)

- 5, - 1, 3, 7 \dots

Here, first term,

a = - 5

Common difference,

d =

Second term

-

First term

d = (- 1) - (- 5) = - 1 + 5 = 4

(iii)

\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{1 3}{3}

Here, first term,

a = \frac{1}{3}

Common difference,

d =

Second term

-

First term

d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}

(iv)

0.6, 1.7, 2.8, 3.9 \dots

Here, first term,

a = 0.6

Common difference,

d =

Second term

-

First term

d = 1.7 - 0.6

= 1.1

Question 4

Which of the following are APs ? If they form an AP, find the common difference

d

and write three more terms.
(i)

2, 4, 8, 16, . . . .

(ii)

2, \frac{5}{2}, 3, \frac{7}{2}, . . .

(iii)

- 1.2, - 3.2, - 5.2, - 7.2, . . .

(iv)

- 10, - 6, - 2, 2, . . .

(v)

3, 3 + 2, 3 + 22, 3 + 32

(vi)

0.2, 0.22, 0.222, 0.2222, . . . .

(vii)

0, - 4, - 8, - 12, . . .

(viii)

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, . . .

(ix)

1, 3, 9, 27

(x)

a, 2 a, 3 a, 4 a, . . .

(xi)

a, a^{2}, a^{3}, a^{4}, . . .

(xii)

2, 8, 18, 32, . . .

(xiii)

3, 6, 9, 12, . . .

(xiv)

1^{2}, 3^{2}, 5^{2}, 7^{2}, . .

(xv)

1^{2}, 5^{2}, 7^{2}, 73, . .

Medium

Solution

Verified by Toppr

a, b, c

are said to be in AP if the common difference between any two consecutive number of the series is same ie

b - a = c - b \Rightarrow 2 b = a + c

(i)

It is not in AP, as the difference between consecutive terms is different.

(i i)

It is in AP with common difference

d = \frac{5}{2} - 2 = \frac{1}{2}

t_{n} = a + (n - 1) d

a = 2

t_{5} = 2 + (5 - 1) \frac{1}{2}

Next three terms are

4, \frac{9}{2}, 5

(i i i)

It is in AP with common difference

d = - 3.2 + 1.2 = - 2

,and

a = - 1.2

Next three terms are

a + (5 - 1) d = - 9.2,

a + (6 - 1) d = - 11.2,

a + (7 - 1) d = - 13.2

(i v)

It is in AP with common difference

d = - 6 + 10 = 4

, and

a = - 10

Next three terms are

a + (5 - 1) d = 6,

a + (6 - 1) d = 10,

a + (7 - 1) d = 14

(v)

It is in AP with common difference

d = 3 + 2 - 3 = 2

, and

a = 3

Next three terms are

a + (5 - 1) d = 3 + 42,

a + (6 - 1) d = 3 + 52,

a + (7 - 1) d = 3 + 62

(v i)

It is not in AP since

0.22 - 0.2 \neq = 0.222 - 0.22

(v i i)

It is in AP with common difference

d = - 4 - 0 = - 4

and

a = 0

Next three terms are

a + (5 - 1) d = - 16,

a + (6 - 1) d = - 20,

a + (7 - 1) d = - 24

(v i i i)

It is in AP, with common difference

0

, therefore next three terms will also be same as previous ones, i.e.,

- \frac{1}{2}

(i x)

It is not in AP since

3 - 1 \neq = 9 - 3

(x)

It is in AP with common difference

d = 2 a - a = a

and first term is

a

Next three terms are

a + (5 - 1) d = 5 a,

a + (6 - 1) d = 6 a,

a + (7 - 1) d = 7 a

(x i)

It is not in AP, as the difference is not constant.

(x i i)

It is in AP with common difference

d = 2

and

a = 2

Next three terms are

a + (5 - 1) d = 52 = 50,

a + (6 - 1) d = 72,

a + (7 - 1) d = 98

(x i i i)

It is not in AP as difference is not constant.

(x i v)

It is not in AP as difference is not constant.

(x v)

It is in AP with common difference

d = 5^{2} - 1 = 24

and

a = 1

Next three terms are

a + (5 - 1) d = 97,

a + (6 - 1) d = 121,

a + (7 - 1) d = 145

Exercise 5.2

Question 1

In the following table, given that

a

is the first term,

d

the common difference and

a_{n}

the

n t h

term of the AP.

	$a$	$d$	$n$	$a_{n}$
(i)	7	3	8	.......
(ii)	-18	.......	10	0
(iii)	.......	-3	18	-5
(iv)	-18.9	2.5	.......	3.6
(v)	3.5	0	105	.......

(i) a_{n} = 13 (i i) d = 1 (i i i) a = 46 (i v) n = 9 (v) a_{n} = 3.5

(i) a_{n} = 36 (i i) d = 7 (i i i) a = 46 (i v) n = 13 (v) a_{n} = 3.5

(i) a_{n} = 18 (i i) d = 3 (i i i) a = 46 (i v) n = 4 (v) a_{n} = 3.5

(i) a_{n} = 28 (i i) d = 2 (i i i) a = 46 (i v) n = 10 (v) a_{n} = 3.5

Medium

Solution

Verified by Toppr

Here we use the formula

t_{n} = a + (n - 1) d

(i).

a_{8} = a + (8 - 1) d = 7 + 7 * 3 = 28

(ii).

a = - 18, d = ?

t_{n} = 1 + (n - 1) d

0 = - 18 + (10 - 1) d

d = 2

(iii).

- 5 = a + (18 - 1) (- 3)

Solve this to get a=46.
(iv).

3.6 = - 18.9 + (n - 1) 2.5

n= 10.
(v).

t_{n} = 3.5 + (105 - 1) 0 = 3.5

Question 2

Choose the correct choice in the following and justify :
(i)

3 0^{t h}

term of the AP:

10, 7, 4, . . .

, is
(A)

97

(B)

77

(C)

- 77

(D)

87

(ii)

1 1^{t h}

term of the AP:

- 3, - \frac{1}{2}, 2

,...., is
(A)

28

(B)

22

(C)

38

(D)

- 48 \frac{1}{2}

Medium

Solution

Verified by Toppr

(i) Given that
A.P.

10, 7, 4, \dots

First term,

a = 10

Common difference,

d = a_{2} = a_{1} = 7 - 10 = - 3

We know that

a_{n} = a + (n - 1) d

a_{30} = 10 + (30 - 1) (- 3)

= 10 + (29) (- 3)

= 10 - 87

= - 77

Hence, the correct answer is option C.

(ii) A.P. is

- 3, \frac{- 1}{2}, 2, . . . . . .

First term

a = - 3

Common difference,

d = a_{2} - a_{1} = \frac{- 1}{2} - (- 3) = \frac{5}{2}

We know that,

a_{n} = a + (n - 1) d

a_{11} = - 3 + (11 - 1) \frac{5}{2} = - 3 + (10) \frac{5}{2} = - 3 + 25 = 22

Hence, the answer is option B.

Question 3

In the following APs, find the missing terms in the boxes :

Medium

Solution

Verified by Toppr

(i) For this A.P.,

a = 2

a_{3} = 26

We know that,

a_{n} = a + (n - 1) d

a_{3} = 2 + (3 - 1) d

26 = 2 + 2 d

24 = 2 d

d = 12

a_{2} = 2 + (2 - 1) 12

= 14

Therefore,

14

is the missing term.

(ii) For this A.P.,

a_{2} = 13

and

a_{4} = 3

We know that,

a_{n} = a + (n - 1) d

a_{2} = a + (2 - 1) d

13 = a + d

... (i)

a_{4} = a + (4 - 1) d

3 = a + 3 d

... (ii)

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

- 10 = 2 d

d = - 5

From equation (i), we get,

13 = a + (- 5)

a = 18

a_{3} = 18 + (3 - 1) (- 5)

= 18 + 2 (- 5) = 18 - 10 = 8

Therefore, the missing terms are

18

and

8

respectively.

(iii) For this A.P.,

a_{1} = 5

and

a_{4} = 9 \frac{1}{2}

We know that,

a_{n} = a + (n - 1) d

a_{4} = 5 + (4 - 1) d

9 \frac{1}{2} = 5 + 3 d

d = \frac{3}{2}

a_{2} = a + d

a_{2} = 5 + \frac{3}{2}

a_{2} = \frac{1 3}{2}

a_{3} = a_{2} + \frac{3}{2}

a_{3} = 8

Therefore, the missing terms are

6 \frac{1}{2}

and

8

respectively.

(iv) For this A.P.,

a = - 4

and

a_{6} = 6

We know that,

a_{n} = a + (n - 1) d

a_{6} = a + (6 - 1) d

6 = - 4 + 5 d

10 = 5 d

d = 2

a_{2} = a + d = - 4 + 2 = - 2

a_{3} = a + 2 d = - 4 + 2 (2) = 0

a_{4} = a + 3 d = - 4 + 3 (2) = 2

a_{5} = a + 4 d = - 4 + 4 (2) = 4

Therefore, the missing terms are

- 2, 0, 2,

and

4

respectively.

(v) For this A.P.,

a_{2} = 38

a_{6} = - 22

We know that

a_{n} = a + (n - 1) d

a_{2} = a + (2 - 1) d

38 = a + d

... (i)

a_{6} = a + (6 - 1) d

- 22 = a + 5 d

... (ii)

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

- 22 - 38 = 4 d

- 60 = 4 d

d = - 15

a = a_{2} - a = 38 - (- 15) = 53

a_{3} = a + 2 d = 53 + 2 (- 15) = 23

a_{4} = a + 3 d = 53 + 3 (- 15) = 8

a_{5} = a + 4 d = 53 + 4 (- 15) = - 7

Therefore, the missing terms are

53, 23, 8

and

- 7

respectively.

Question 4

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

Easy

Solution

Verified by Toppr

The given sequence is 3, 8, 13, 18, ....78.

'a' = 3, 'd' = 5, last term = 78, 'n' = ?

t_{n} = a + (n - 1) d ⟹ 78 = 3 + (n - 1) 5 ⟹ 78 = 3 + 5 n - 5 ⟹ 78 = 5 n - 2 ⟹ 5 n = 80 o r n = 16

.

The 16th term is 78.

Question 5

Find the number of terms in the following

A . P . 7, 13, 19, . . . ., 205

Easy

Solution

Verified by Toppr

Let

a, d

and

l

be the first term, common difference and the last term of the A.P. respectively.

Given

a = 7, d = 13 - 7 = 6

and

l = 205

Since

l = a + (n - 1) d

\Rightarrow 205 = 7 + (n - 1) 6

\Rightarrow 198 = (n - 1) 6

\Rightarrow 33 = n - 1

\Rightarrow n = 34

Hence, there are

34

terms in the A.P.

Question 6

11, 8, 5, 2, . . . .

In this

A . P .

which term is number

- 151

Medium

Solution

Verified by Toppr

D = - 3

a + (n - 1) d = - 151

11 + (n - 1) - 3 = - 151

11 - 3 n + 3 = - 151

- 3 n = - 165

n = 55

55 t h t e r m

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NCERT Solutions for Class 10 Maths Chapter 5 : Arithmetic Progressions

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions – Brief Overview

1. Arithmetic Progressions

Except for the initial term, an arithmetic progression is a list of integers in which each item is obtained by adding a fixed number to the prior term. This set figure is known as the AP's common difference, which might be positive, negative, or zero.

Arithmetic Progression, abbreviated as AP, is a number series. This sequence exists in such a way that the difference between any two subsequent numbers is constant.

Any arithmetic progression can be expressed as: a, a + d, a + 2d, a + 3d,…

where a is the initial term and d is the common difference.

2. The formula for common difference (d)

A number sequence a1, a2, a3.... is an AP if the difference a₂ – a₁, a₃ – a₂, a₄ – a₃, .... equals the same value.

d = a₁ – a_n-1

3. The n^th term of an Arithmetic Progression

This formula can be used to obtain the value of the nth term of an arithmetic progression in Class 10 Maths Chapter 5 NCERT solutions. The n^th term of an Arithmetic Progression with the first term ‘a’ and a common difference ‘d’ is given as:

a_n = a + (n-1)d

4. Sum of the first n terms of an Arithmetic Progression

If the value of the first term and the total terms are known for an arithmetic progression, the sum of the first n terms can be calculated.

The sum of an A.P.'s first n terms is given by:

The Sum of the first n positive integers is given by:

If l is the last term of a finite A.P., then the sum of all terms of the A.P. is given by:

5. The Arithmetic Mean of Two Numbers

If a, b, and c are all in AP. The arithmetic mean of a and c is then b, which is given by:

b = (a + c) / 2

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles

Chapter 13 : Surface Areas and Volumes

Chapter 14 : Statistics

Chapter 15 : Probability

Frequently Asked Questions on NCERT Class 10 Maths Chapter 5 : Arithmetic Progressions

Q1. How does one calculate the sum of an arithmetic progression?

Answer: If the value of the first term, the common difference and the total terms are known for an arithmetic progression, then the sum of the first n terms can be calculated.

The sum of an A.P.'s first n terms is given by:

Q2. How many exercises are there in Chapter 5 of Class 10 Maths?

Answer: This chapter has four exercises with examples that are comparable. The expert-curated class 10 mathematics chapter 5 Arithmetic Progression consists of 49 questions. There are 20 long answers, 20 moderate questions, and 9 easy questions among the 49 issues. Multiple choice questions, descriptive type questions, long answer type questions, short answer type questions, fill in the gaps, and daily life examples are included in Chapter 5 of NCERT Solutions for Class 10 Maths.

Q3. Is it necessary to practice all of the questions in NCERT Solutions Class 10 Maths Arithmetic Progressions?

Answer: The NCERT Solutions Class 10 Maths Chapter 5 has carefully curated real-life examples and exercise questions that will assist students in delving into the concept of arithmetic progression and understanding the fundamental basic terminologies and facts. Math takes constant practice to imprint the fundamental principles in mind for a long time. As a result, it would be incredibly advantageous for students to practice all of the questions in order to build a solid foundation for the chapter while also improving their logical and reasoning skills.

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 11 : Constructions

Chapter 12 : Area Related to Circles