Arithmetic Progressions Class 10 Maths Chapter 5 Exercise 5.1 PDF
Neha Tanna23 Nov, 2025
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NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions: Chapter 5 of Class 10 Maths introduces Arithmetic Progressions (APs), where each term changes by a fixed value. Exercise 5.1 helps students identify whether a sequence is an AP and find its common difference d.
This exercise builds strong pattern-recognition and reasoning skills, preparing learners for more complex AP problems ahead. These concepts are especially helpful for students studying Arithmetic Progressions class 10 exercise 5.1.
Class 10 Arithmetic Progressions Exercise 5.1 Solutions
Below are the NCERT solutions for Class 10 Maths Chapter 5 Exercise 5.1 Arithmetic Progressions, providing clear explanations to help students understand the basics of AP, especially useful for practice.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 = 15Taxi 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 1/4 of the air remaining in the cylinder at a time.
Solution:
Let the volume of air in a cylinder initially be V 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 , 3 V /4 , (3 V /4) 2 , (3 V /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.150Cost 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) nTherefore, 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 = 10Let us consider, the Arithmetic Progression series be a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = 10 a 2 = a 1 + d = 10+10 = 20 a 3 = a 2 + d = 20+10 = 30 a 4 = a 3 + d = 30+10 = 40 a 5 = a 4 + 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 a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = -2 a 2 = a 1 + d = – 2+0 = – 2 a 3 = a 2 +d = – 2+0 = – 2 a 4 = a 3 + 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 a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = 4 a 2 = a 1 + d = 4-3 = 1 a 3 = a 2 + d = 1-3 = – 2 a 4 = a 3 + 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 a 1 , a 2 , a 3 , a 4 , a 5 … a 2 = a 1 + d = -1+1/2 = -1/2 a 3 = a 2 + d = -1/2+1/2 = 0 a 4 = a 3 + 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 a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = a = – 1.25 a 2 = a 1 + d = – 1.25-0.25 = – 1.50 a 3 = a 2 + d = – 1.50-0.25 = – 1.75 a 4 = a 3 + 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
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 ⇒ 1.14. 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 , 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 , 7 3 …
Solution
Here, the common difference is a 2 – a 1 = 4 – 2 = 2 a 3 – a 2 = 8 – 4 = 4 a 4 – a 3 = 16 – 8 = 8 a n +1 – a n 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, a 2 – a 1 = 5/2-2 = 1/2 a 3 – a 2 = 3-5/2 = 1/2 a 4 – a 3 = 7/2-3 = 1/2 a n +1 – a n 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 a 5 = 7/2+1/2 = 4 a 6 = 4 +1/2 = 9/2 a 7 = 9/2 +1/2 = 5(iii) Given, -1.2, – 3.2, -5.2, -7.2 …
Here, a 2 – a 1 = (-3.2)-(-1.2) = -2 a 3 – a 2 = (-5.2)-(-3.2) = -2 a 4 – a 3 = (-7.2)-(-5.2) = -2 a n +1 – a n 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 a 5 = – 7.2-2 = -9.2 a 6 = – 9.2-2 = – 11.2 a 7 = – 11.2-2 = – 13.2(iv) Given, -10, – 6, – 2, 2 …
Here, the terms and their difference are a 2 – a 1 = (-6)-(-10) = 4 a 3 – a 2 = (-2)-(-6) = 4 a 4 – a 3 = (2 -(-2) = 4 a n +1 – a n 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 a 5 = 2+4 = 6 a 6 = 6+4 = 10 a 7 = 10+4 = 14(v) Given, 3, 3+√2, 3+2√2, 3+3√2
Here, a 2 – a 1 = 3+√2-3 = √2 a 3 – a 2 = (3+2√2)-(3+√2) = √2 a 4 – a 3 = (3+3√2) – (3+2√2) = √2 a n +1 – a n 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 a 5 = (3+√2) +√2 = 3+4√2 a 6 = (3+4√2)+√2 = 3+5√2 a 7 = (3+5√2)+√2 = 3+6√2(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here, a 2 – a 1 = 0.22-0.2 = 0.02 a 3 – a 2 = 0.222-0.22 = 0.002 a 4 – a 3 = 0.2222-0.222 = 0.0002 a n +1 – a n 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, a 2 – a 1 = (-4)-0 = -4 a 3 – a 2 = (-8)-(-4) = -4 a 4 – a 3 = (-12)-(-8) = -4 a n +1 – a n 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 a 5 = -12-4 = -16 a 6 = -16-4 = -20 a 7 = -20-4 = -24(viii) -1/2, -1/2, -1/2, -1/2 ….
Here, a 2 – a 1 = (-1/2) – (-1/2) = 0 a 3 – a 2 = (-1/2) – (-1/2) = 0 a 4 – a 3 = (-1/2) – (-1/2) = 0 a n +1 – a n 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; a 5 = (-1/2)-0 = -1/2 a 6 = (-1/2)-0 = -1/2 a 7 = (-1/2)-0 = -1/2(ix) 1, 3, 9, 27 …
Here, a 2 – a 1 = 3-1 = 2 a 3 – a 2 = 9-3 = 6 a 4 – a 3 = 27-9 = 18 a n +1 – a n or the common difference is not the same every time. Therefore, the given series doesn’t form an A.P.
(x) a , 2 a , 3 a , 4 a …
(xi) a , a 2 , a 3 , a 4 …
Here, a 2 – a 1 = a 2 – a = a( a -1) a 3 – a 2 = a 3 – a 2 = a 2 ( a -1) a 4 – a 3 = a 4 – a 3 = a 3 ( a -1) a n +1 – a n 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 …
Therefore,d = √2, and the given series forms an A.P. Hence, the next three terms are a 5 = √32+√2 = 4√2+√2 = 5√2 = √50 a 6 = 5√2+√2 = 6√2 = √72 a 7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
Therefore, the given series doesn’t form an A.P.
(xiv) 1 2 , 3 2 , 5 2 , 7 2 …
Therefore, the given series doesn’t form an A.P.
(xv) 1 2 , 5 2 , 7 2 , 73 …
Therefore, d = 24 and the given series forms an A.P. Hence, the next three terms are a 5 = 73+24 = 97 a 6 = 97+24 = 121 a 7 = 121+24 = 145
NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 PDF
Exercise 5.1 explains how to identify an AP and compute its common difference, building key skills for upcoming concepts.
To make learning easier, a detailed PDF with step-by-step solutions has been provided, helping students revise effectively for exams. This resource is ideal for those looking for Class 10 Arithmetic Progressions Exercise 5.1.
NCERT solutions Class 10 Maths PDF