The exercise solutions provided here are the result of thorough research and experience of subject experts at BYJU’S. This NCERT Solutions page has answers to all the questions provided in the NCERT Class 10 Maths textbook. Solving this exercise will help students tune in with the introduction and basic concepts involved in Arithmetic progression.
This NCERT solution page will contain the methodology to solve the exercise questions 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.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1
Access other exercise solutions of Class 10 Maths Chapter 5 – Arithmetic Progression
The following are the exercises covered in NCERT Class 10 Solutions of Maths Chapter 5.
Exercise 5.2– 20 questions (1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions)
Exercise 5.3– 20 Questions (3 fill in the blanks, 4 daily life examples, and 13 descriptive-type questions)
Exercise 5.4– 5 Questions (5 Long answer questions)
Access answers of Maths NCERT Class 10 Chapter 5 – Arithmetic Progression Exercise 5.1
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.
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 on……
Thus, 15, 23, 31, 39 … 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 be VÂ litres.
In each stroke, the vacuum pump removes 1/4th of the remaining air in the cylinder at a time. Or we can say after every stroke, 1-1/4 = 3/4th part of the air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3…and so on
Clearly, we can see here the adjacent terms of this series do not have a 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 … 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 10,000 is deposited at compound interest at 8% per annum.
Solution:
We know that if Rs. P is deposited at 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)3……
Clearly, the terms of this series do not have a common difference between them. Therefore, this is not an A.P.
2. Write the first four terms of the A.P. when the first term a, and the common difference are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = – 3
(iv) a = -1 d = 1/2
(v) a = – 1.25, d = – 0.25
Solutions:
(i) a = 10, d = 10
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+d = 40+10 = 50
And so on…
Therefore, the A.P. series will be 10, 20, 30, 40, 50 …
And the first four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3Â =Â a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …
And the first four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore, the A.P. series will be 4, 1, – 2 – 5 …
And the first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 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) a = – 1.25, d = – 0.25
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = – 1.25
a2 = a1 + d = – 1.25-0.25 = – 1.50
a3 = a2 + d = – 1.50-0.25 = – 1.75
a4 = a3 + d = – 1.75-0.25 = – 2.00
Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
And the first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …
Solutions
(i) Given series,
3, 1, – 1, – 3 …
The first term, a = 3
The common difference, d = Second term – First term
⇒ 1 – 3 = -2
⇒ d = -2
(ii) Given series, – 5, – 1, 3, 7 …
The first term, a = -5
The common difference, d = Second term – First term
⇒ ( – 1)-( – 5) = – 1+5 = 4
(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….
The first term, a = 1/3
The common difference, d = Second term – First term
⇒ 5/3 – 1/3 = 4/3
(iv) Given series, 0.6, 1.7, 2.8, 3.9 …
The first term, a = 0.6
The common difference, d = Second term – First term
⇒ 1.7 – 0.6
⇒ 1.1
4. Which of the following are APs? If they form an A.P., find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a2, a3, a4 …
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72 …
(xv) 12, 52, 72, 73 …
Solution
(i) Given to us,
2, 4, 8, 16 …
Here, the common difference is
a2Â –Â a1Â = 4 – 2 = 2
a3Â –Â a2Â = 8 – 4 = 4
a4Â –Â a3Â = 16 – 8 = 8
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series is not forming an A.P.
(ii) Given, 2, 5/2, 3, 7/2 ….
Here,
a2Â –Â a1Â =Â 5/2-2 = 1/2
a3Â –Â a2Â =Â 3-5/2 = 1/2
a4Â –Â a3Â =Â 7/2-3 = 1/2
an+1Â –Â an or the common difference is the same every time.
Therefore, d = 1/2 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) Given, -1.2, – 3.2, -5.2, -7.2 …
Here,
a2Â –Â a1Â = (-3.2)-(-1.2) = -2
a3Â –Â a2Â = (-5.2)-(-3.2) = -2
a4Â –Â a3Â = (-7.2)-(-5.2) = -2
an+1Â –Â an or common difference is the same every time.
Therefore, d = -2 and the given series are in A.P.
Hence, the next three terms are
a5Â = – 7.2-2 = -9.2
a6Â = – 9.2-2 = – 11.2
a7Â = – 11.2-2 = – 13.2
(iv) Given, -10, – 6, – 2, 2 …
Here, the terms and their difference are
a2Â –Â a1Â = (-6)-(-10) = 4
a3Â –Â a2Â = (-2)-(-6) = 4
a4Â –Â a3Â = (2 -(-2) = 4
an+1Â –Â an or the common difference is the same every time.
Therefore, d = 4, and the given numbers are in A.P.
Hence, the next three terms are
a5Â = 2+4 = 6
a6Â = 6+4 = 10
a7Â = 10+4 = 14
(v) Given, 3, 3+√2, 3+2√2, 3+3√2
Here,
a2 – a1 = 3+√2-3 = √2
a3 – a2 = (3+2√2)-(3+√2) = √2
a4 – a3 = (3+3√2) – (3+2√2) = √2
an+1Â –Â an or the common difference is the same every time.
Therefore, d = √2, and the given series forms an A.P.
Hence, the next three terms are
a5 = (3+√2) +√2 = 3+4√2
a6 = (3+4√2)+√2 = 3+5√2
a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2Â –Â a1Â =Â 0.22-0.2 = 0.02
a3Â –Â a2Â =Â 0.222-0.22 = 0.002
a4Â –Â a3Â =Â 0.2222-0.222 = 0.0002
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series doesn’t form an A.P.
(vii) 0, -4, -8, -12 …
Here,
a2Â –Â a1Â =Â (-4)-0 = -4
a3Â –Â a2Â =Â (-8)-(-4) = -4
a4Â –Â a3Â =Â (-12)-(-8) = -4
an+1Â –Â an or the common difference is the same every time.
Therefore, d = -4 and the given series forms an A.P.
Hence, the 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 ….
Here,
a2Â –Â a1Â = (-1/2) – (-1/2) = 0
a3Â –Â a2Â = (-1/2) – (-1/2) = 0
a4Â –Â a3Â = (-1/2) – (-1/2) = 0
an+1Â –Â an or the common difference is the same every time.
Therefore, d = 0 and the given series forms an A.P.
Hence, the 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 …
Here,
a2Â –Â a1Â =Â 3-1 = 2
a3Â –Â a2Â =Â 9-3 = 6
a4Â –Â a3Â =Â 27-9 = 18
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series doesn’t form an A.P.
(x) a, 2a, 3a, 4a …
Here,
a2 – a1 = 2a–a = a
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
an+1Â –Â an or the common difference is the same every time.
Therefore, d = a and the given series forms an A.P.
Hence, the next three terms are
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(xi) a, a2, a3, a4 …
Here,
a2 – a1 = a2–a = a(a-1)
a3Â –Â a2Â =Â a3Â –Â a2Â =Â a2(a-1)
a4Â –Â a3Â =Â a4Â –Â a3Â =Â a3(a-1)
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series doesn’t form an A.P.
(xii) √2, √8, √18, √32 …
Here,
a2 – a1 = √8-√2  = 2√2-√2 = √2
a3 – a2 = √18-√8 = 3√2-2√2 = √2
a4 – a3 = 4√2-3√2 = √2
an+1Â –Â an or the common difference is the same every time.
Therefore, d = √2, and the given series forms an A.P.
Hence, the next three terms are
a5 = √32+√2 = 4√2+√2 = 5√2 = √50
a6 = 5√2+√2 = 6√2 = √72
a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
Here,
a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series doesn’t form an A.P.
(xiv) 12, 32, 52, 72Â …
Or, 1, 9, 25, 49 …..
Here,
a2 − a1 = 9−1 = 8
a3 − a2 = 25−9 = 16
a4 − a3 = 49−25 = 24
an+1Â –Â an or the common difference is not the same every time.
Therefore, the given series doesn’t form an A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2 − a1 = 25−1 = 24
a3 − a2 = 49−25 = 24
a4 − a3 = 73−49 = 24
an+1Â –Â an or the common difference is the same every time.
Therefore, d = 24 and the given series forms an A.P.
Hence, the next three terms are
a5Â = 73+24 = 97
a6Â = 97+24 = 121
a7Â = 121+24 = 145
Topics covered in Exercise 5.1
In the NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1, the following topics are discussed.
- Introduction
- Basics of arithmetic progression
How Is It Helpful?
- Provides answers to questions of Exercise 5.1
- NCERT Solutions Class 10 Maths helps students understand basic concepts of arithmetic progression
- After studying this exercise page, learners will be able to solve less complex questions on arithmetic progression.
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