Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.1 introduces the concept of Arithmetic Progressions (AP), where numbers follow a fixed pattern based on a constant difference. As part of the CBSE Class 10th syllabus, this exercise will help you recognise whether a sequence forms an AP and determine its common difference.

The step-by-step NCERT solutions make it easier to follow how patterns are identified and verified. Practising these questions builds a strong base for understanding formulas and solving more advanced problems in later exercises.

NCERT Solutions for Class 10 Maths Chapter 5 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 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 1/4 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.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  a 1 , a 2 , a 3 , a 4 , a 5 … a 1 = 

a = 10 

a 2 = a 1 + d = 10+10 = 20 

a3 = 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 to be a1 , 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 to 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

(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 , 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

(i) Given to us, 2, 4, 8, 16 …

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 …

Here, a 2 – a 1 = 2 a – a = a a 3 – a 2 = 3 a -2 a = a a 4 – a 3 = 4 a -3 a = a a n +1 – a n 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 a 5 = 4 a + a = 5 a a 6 = 5 a + a = 6 a a 7 = 6 a + a = 7 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 …

Here, a 2 – a 1 = √8-√2  = 2√2-√2 = √2 a 3 – a 2 = √18-√8 = 3√2-2√2 = √2 a 4 – a 3 = 4√2-3√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 = √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 …

Here, a 2 – a 1 = √6-√3 = √3×√2-√3 = √3(√2-1) a 3 – a 2 = √9-√6 = 3-√6 = √3(√3-√2) a 4 – a 3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3) 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.

(xiv) 1 2 , 3 2 , 5 2 , 7 2 …

Or, 1, 9, 25, 49 ….. Here, a 2 − a 1 = 9−1 = 8 a 3 − a 2 = 25−9 = 16 a 4 − a 3 = 49−25 = 24 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.

(xv) 1 2 , 5 2 , 7 2 , 73 …

Or 1, 25, 49, 73 … Here, a 2 − a 1 = 25−1 = 24 a 3 − a 2 = 49−25 = 24 a 4 − a 3 = 73−49 = 24 a n +1 – a n 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 a 5 = 73+24 = 97 a 6 = 97+24 = 121 a 7 = 121+24 = 145

How to Score Better in Class 10 Maths Exam?

Scoring well in Class 10 Maths requires clear concepts, regular practice, and a focus on accuracy and answer presentation. To score better, you should:

• Build Strong Concepts:

Focus on understanding concepts in Class 10 Maths instead of memorising steps, as this helps in solving application-based questions.

• Work on Weak Areas:

Focus on difficult topics from the Class 10 Maths syllabus instead of skipping them to avoid losing marks.

• Revise Formulas Daily:

Regular revision of PW Class 10 Maths MIQs helps avoid calculation mistakes in exams.

• Practise Regularly:

Solve all CBSE Class 10 NCERT questions multiple times to strengthen your basics and improve accuracy.

• Solve Previous Year Papers:

Practising CBSE Class 10 Maths previous year questions (PYQs) helps you understand question patterns and important topics.