# NCERT Solutions for Class 10 Maths Exercise 5.1 Chapter 5 Arithmetic Progressions And Geometric Progression

The exercise solutions provided here are the result of thorough research and experience of subject experts, at BYJUâ€™S. This NCERT Solution page has answers to all the questions provided in the NCERT Class 10 Maths textbook. Solving this exercise will make you tuned-in with introduction and basic concepts involved in Arithmetic progression.

This NCERT solution page will contain the methodology to solve the exercise question provided on page 99 of NCERT class 10 Maths textbook. After studying this solution, you will be able to solve simple questions on Arithmetic progressions.

### Access other exercise solutions of Class 10 Maths Chapter 5- Arithmetic Progression

Exercise 5.2â€“ 20 questions (1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions)
Exercise 5.3â€“ 20 Questions (3 fill in the blanks, 4 daily life examples, and 13 descriptive type questions)
Exercise 5.4â€“ 5 Questions (5 Long answer questions)

### Access Answers of Maths NCERT Class 10 Chapter 5 â€“ Arithmetic Progression Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as 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.

Solution:

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so onâ€¦â€¦

Thus, 15, 23, 31, 39 â€¦ forms an A.P. because every next term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4Â of the air remaining in the cylinder at a time.

Solution:

Let the volume of air in a cylinder, initially, beÂ VÂ litres.

In each stroke, the vacuum pump removesÂ 1/4thÂ of airÂ remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will beÂ V, 3V/4 , (3V/4)2Â , (3V/4)3â€¦and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

(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.

Solution:

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 â€¦ forms an A.P. with a common difference of 50 between each term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution:

We know that if Rs. PÂ is deposited atÂ r% compound interest per annum for n years, the amount of money will be:

P(1+r/100)n

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100)2, 10000(1+8/100)3â€¦â€¦

Clearly, the terms of this series do not have the common difference between them. Therefore, this isÂ not an A.P.

2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:

(i)Â aÂ = 10,Â dÂ = 10
(ii)Â aÂ = -2,Â dÂ = 0
(iii)Â aÂ = 4,Â dÂ = â€“ 3
(iv)Â aÂ = -1Â dÂ = 1/2
(v)Â aÂ = â€“ 1.25,Â dÂ = â€“ 0.25

Solutions:

(i)Â aÂ = 10,Â dÂ = 10

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = 10

a2Â =Â a1+dÂ = 10+10 = 20

a3Â =Â a2+dÂ = 20+10 = 30

a4Â =Â a3+dÂ = 30+10 = 40

a5Â =Â a4+dÂ = 40+10 = 50

And so onâ€¦

Therefore, the A.P. series will be 10, 20, 30, 40, 50 â€¦

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii)Â aÂ = â€“ 2,Â dÂ = 0

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = -2

a2Â =Â a1+dÂ = â€“ 2+0 = â€“ 2

a3Â =Â a2+d = â€“ 2+0 = â€“ 2

a4Â =Â a3+dÂ = â€“ 2+0 = â€“ 2

Therefore, the A.P. series will be â€“ 2, â€“ 2, â€“ 2, â€“ 2 â€¦

And, First four terms of this A.P. will be â€“ 2, â€“ 2, â€“ 2 and â€“ 2.

(iii)Â aÂ = 4,Â dÂ = â€“ 3

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = 4

a2Â =Â a1+dÂ = 4-3 = 1

a3Â =Â a2+dÂ = 1-3 = â€“ 2

a4Â =Â a3+dÂ = -2-3 = â€“ 5

Therefore, the A.P. series will be 4, 1, â€“ 2 â€“ 5 â€¦

And, first four terms of this A.P. will be 4, 1, â€“ 2 and â€“ 5.

(iv)Â aÂ = â€“ 1,Â dÂ = 1/2

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a2Â =Â a1+dÂ = -1+1/2 = -1/2

a3Â =Â a2+dÂ = -1/2+1/2 = 0

a4Â =Â a3+dÂ = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v)Â aÂ = â€“ 1.25,Â dÂ = â€“ 0.25

Let us consider, the Arithmetic Progression series beÂ a1,Â a2,Â a3,Â a4,Â a5Â â€¦

a1Â =Â aÂ = â€“ 1.25

a2Â =Â a1Â +Â dÂ = â€“ 1.25-0.25 = â€“ 1.50

a3Â =Â a2Â +Â dÂ = â€“ 1.50-0.25 = â€“ 1.75

a4Â =Â a3Â +Â dÂ = â€“ 1.75-0.25 = â€“ 2.00

Therefore, the A.P series will be 1.25, â€“ 1.50, â€“ 1.75, â€“ 2.00 â€¦â€¦..

And first four terms of this A.P. will be â€“ 1.25, â€“ 1.50, â€“ 1.75 and â€“ 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, â€“ 1, â€“ 3 â€¦
(ii) -5, â€“ 1, 3, 7 â€¦
(iii) 1/3, 5/3, 9/3, 13/3 â€¦.
(iv) 0.6, 1.7, 2.8, 3.9 â€¦

Solutions

(i) Given series,

3, 1, â€“ 1, â€“ 3 â€¦

First term,Â aÂ = 3

Common difference,Â dÂ = Second term â€“ First term

â‡’Â  1 â€“ 3 = -2

â‡’Â  d = -2

(ii) Given series, â€“ 5, â€“ 1, 3, 7 â€¦

First term,Â aÂ = -5

Common difference,Â dÂ = Second term â€“ First term

â‡’Â ( â€“ 1)-( â€“ 5) = â€“ 1+5 = 4

(iii) Given series, 1/3, 5/3, 9/3, 13/3 â€¦.

First term,Â aÂ = 1/3

Common difference,Â dÂ = Second term â€“ First term

â‡’Â 5/3 â€“ 1/3 = 4/3

(iv) Given series, 0.6, 1.7, 2.8, 3.9 â€¦

First term,Â aÂ = 0.6

Common difference,Â dÂ = Second term â€“ First term

â‡’Â 1.7 â€“ 0.6

â‡’Â 1.1

4. Which of the following are APs? If they form an A.P. find the common differenceÂ dÂ and write three more terms.

(i) 2, 4, 8, 16 â€¦
(ii) 2, 5/2, 3, 7/2 â€¦.
(iii) -1.2, -3.2, -5.2, -7.2 â€¦
(iv) -10, â€“ 6, â€“ 2, 2 â€¦
(v) 3, 3 +Â âˆš2, 3Â + 2âˆš2, 3Â + 3âˆš2
(vi) 0.2, 0.22, 0.222, 0.2222 â€¦.
(vii) 0, â€“ 4, â€“ 8, â€“ 12 â€¦
(viii) -1/2, -1/2, -1/2, -1/2 â€¦.
(ix) 1, 3, 9, 27 â€¦
(x)Â a, 2a, 3a, 4aÂ â€¦
(xi)Â a,Â a2,Â a3,Â a4Â â€¦
(xii) âˆš2, âˆš8, âˆš18, âˆš32Â â€¦
(xiii) âˆš3, âˆš6, âˆš9, âˆš12Â â€¦
(xiv) 12, 32, 52, 72Â â€¦
(xv) 12, 52, 72, 73Â â€¦

Solution

(i) Given to us,

2, 4, 8, 16 â€¦

Here, the common difference is;

a2Â â€“Â a1Â = 4 â€“ 2 = 2

a3Â â€“Â a2Â = 8 â€“ 4 = 4

a4Â â€“Â a3Â = 16 â€“ 8 = 8

Since,Â an+1Â â€“Â an or the common difference is not the same every time.

Therefore, the given series are not forming an A.P.

(ii) Given, 2, 5/2, 3, 7/2 â€¦.

Here,

a2Â â€“Â a1Â =Â 5/2-2 = 1/2

a3Â â€“Â a2Â =Â 3-5/2 = 1/2

a4Â â€“Â a3Â =Â 7/2-3 = 1/2

Since, an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 1/2Â and the given series are in A.P.

The next three terms are;

a5Â = 7/2+1/2 = 4

a6Â = 4Â +1/2 = 9/2

a7Â = 9/2Â +1/2 = 5

(iii)Â Given, -1.2, â€“ 3.2, -5.2, -7.2 â€¦

Here,

a2Â â€“Â a1Â = (-3.2)-(-1.2) = -2

a3Â â€“Â a2Â = (-5.2)-(-3.2) = -2

a4Â â€“Â a3Â = (-7.2)-(-5.2) = -2

Since, an+1Â â€“Â anÂ or common difference is same every time.

Therefore,Â dÂ = -2 and the given series are in A.P.

Hence, next three terms are;

a5Â = â€“ 7.2-2 = -9.2

a6Â = â€“ 9.2-2 = â€“ 11.2

a7Â = â€“ 11.2-2 = â€“ 13.2

(iv) Given, -10, â€“ 6, â€“ 2, 2 â€¦

Here, the terms and their difference are;

a2Â â€“Â a1Â = (-6)-(-10) = 4

a3Â â€“Â a2Â = (-2)-(-6) = 4

a4Â â€“Â a3Â = (2 -(-2) = 4

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 4 and the given numbers are in A.P.

Hence, next three terms are;

a5Â = 2+4 = 6

a6Â = 6+4 = 10

a7Â = 10+4 = 14

(v) Given, 3, 3+âˆš2, 3+2âˆš2, 3+3âˆš2

Here,

a2Â â€“Â a1Â = 3+âˆš2-3 = âˆš2

a3Â â€“Â a2Â = (3+2âˆš2)-(3+âˆš2) = âˆš2

a4Â â€“Â a3Â = (3+3âˆš2) â€“ (3+2âˆš2) = âˆš2

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â âˆš2Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = (3+âˆš2)Â +âˆš2Â = 3+4âˆš2

a6Â = (3+4âˆš2)+âˆš2Â = 3+5âˆš2

a7Â = (3+5âˆš2)+âˆš2Â = 3+6âˆš2

(vi)Â 0.2, 0.22, 0.222, 0.2222 â€¦.

Here,

a2Â â€“Â a1Â =Â 0.22-0.2 = 0.02

a3Â â€“Â a2Â =Â 0.222-0.22 = 0.002

a4Â â€“Â a3Â =Â 0.2222-0.222 = 0.0002

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, and the given series doesnâ€™t forms a A.P.

(vii)Â 0, -4, -8, -12 â€¦

Here,

a2Â â€“Â a1Â =Â (-4)-0 = -4

a3Â â€“Â a2Â =Â (-8)-(-4) = -4

a4Â â€“Â a3Â =Â (-12)-(-8) = -4

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â -4Â and the given series forms a A.P.

Hence, next three terms are;

a5Â =Â -12-4 = -16

a6Â =Â -16-4 = -20

a7Â =Â -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 â€¦.

Here,

a2Â â€“Â a1Â = (-1/2) â€“ (-1/2) = 0

a3Â â€“Â a2Â = (-1/2) â€“ (-1/2) = 0

a4Â â€“Â a3Â = (-1/2) â€“ (-1/2) = 0

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 0 and the given series forms a A.P.

Hence, next three terms are;

a5Â = (-1/2)-0 = -1/2

a6Â =Â (-1/2)-0 = -1/2

a7Â =Â (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 â€¦

Here,

a2Â â€“Â a1Â =Â 3-1 = 2

a3Â â€“Â a2Â =Â 9-3 = 6

a4Â â€“Â a3Â =Â 27-9 = 18

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, and the given series doesnâ€™t form a A.P.

(x)Â a, 2a, 3a, 4aÂ â€¦

Here,

a2Â â€“Â a1Â =Â 2aâ€“aÂ =Â a

a3Â â€“Â a2Â =Â 3a-2aÂ =Â a

a4Â â€“Â a3Â =Â 4a-3aÂ =Â a

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â aÂ and the given series forms a A.P.

Hence, next three terms are;

a5Â =Â 4a+aÂ = 5a

a6Â = 5a+aÂ = 6a

a7Â =Â 6a+aÂ = 7a

(xi)Â a,Â a2,Â a3,Â a4Â â€¦

Here,

a2Â â€“Â a1Â =Â a2â€“aÂ = a(a-1)

a3Â â€“Â a2Â =Â a3Â â€“Â a2Â =Â a2(a-1)

a4Â â€“Â a3Â =Â a4Â â€“Â a3Â =Â a3(a-1)

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t forms a A.P.

(xii) âˆš2, âˆš8, âˆš18, âˆš32Â â€¦

Here,

a2Â â€“Â a1Â = âˆš8-âˆš2Â Â = 2âˆš2-âˆš2Â = âˆš2

a3Â â€“Â a2Â = âˆš18-âˆš8Â = 3âˆš2-2âˆš2Â = âˆš2

a4Â â€“Â a3Â = 4âˆš2-3âˆš2Â = âˆš2

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ =Â âˆš2Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = âˆš32+âˆš2Â = 4âˆš2+âˆš2Â = 5âˆš2Â = âˆš50

a6Â = 5âˆš2+âˆš2Â = 6âˆš2Â = âˆš72

a7Â = 6âˆš2+âˆš2Â = 7âˆš2Â = âˆš98

(xiii)Â âˆš3, âˆš6, âˆš9, âˆš12Â â€¦

Here,

a2Â â€“Â a1Â = âˆš6-âˆš3Â = âˆš3Ã—âˆš2-âˆš3Â = âˆš3(âˆš2-1)

a3Â â€“Â a2Â = âˆš9-âˆš6Â = 3-âˆš6Â = âˆš3(âˆš3-âˆš2)

a4Â â€“Â a3Â = âˆš12Â â€“ âˆš9Â = 2âˆš3Â â€“ âˆš3Ã—âˆš3 = âˆš3(2-âˆš3)

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t form a A.P.

(xiv) 12, 32, 52, 72Â â€¦

Or, 1, 9, 25, 49 â€¦..

Here,

a2Â âˆ’Â a1Â = 9âˆ’1 = 8

a3Â âˆ’Â a2Â = 25âˆ’9 = 16

a4Â âˆ’Â a3Â = 49âˆ’25 = 24

Since,Â an+1Â â€“Â anÂ or the common difference is not same every time.

Therefore, the given series doesnâ€™t form a A.P.

(xv) 12, 52, 72, 73 â€¦

Or 1, 25, 49, 73 â€¦

Here,

a2Â âˆ’Â a1Â = 25âˆ’1 = 24

a3Â âˆ’Â a2Â = 49âˆ’25 = 24

a4Â âˆ’Â a3Â = 73âˆ’49 = 24

Since,Â an+1Â â€“Â anÂ or the common difference is same every time.

Therefore,Â dÂ = 24Â and the given series forms a A.P.

Hence, next three terms are;

a5Â = 73+24 = 97

a6Â = 97+24 = 121

a7Â = 121+24 = 145

#### Topics covered in Exercise 5.1

1. Introduction
2. Basics of arithmetic progression

### How it is helpful

• Provides answers to questions of Exercise 5.1
• Helps you in understanding basic concepts of arithmetic progression
• After studying this exercise page, you will be able to solve less complex questions on arithmetic progression.