Class 10 Maths Chapter 5: Arithmetic Progressions NCERT Solutions

Overview of Class 10 Maths Chapter 5: Arithmetic Progressions

In the NCERT Solutions for Class 10 Maths Chapter 5, students get to learn about Arithmetic Progressions (AP). Thus, this chapter is going to be one of the most important chapters in the course as it will not only allow them to know better about the numbers but also help them to solve related ones quite easily.

The Class 10 Maths Chapter 5 PDF is the one that is best to use and hardest to get for free. Students can download this PDF from ToppersSky App to not only get familiar with the exam but also to score good marks in it.

Deploying the services of seasoned academic professionals, the solutions are designed strictly with the latest CBSE guidelines in mind. The entire question to answer process is described in such a way that it becomes very clear to the student what method to use and how to write the answer in the exam correctly.

Important Points from NCERT Class 10 Arithmetic Progressions

• An arithmetic progression (AP) is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference d, where
  d = a2 − a1.
• The first term of an AP is denoted by a and the nth term by an.
• You can find the nth term in the AP using the formula:
  an = a + (n − 1)d
• The sum of the first n terms Sn in an AP is calculated using the formula:
  Sn = n⁄2 [2a + (n − 1)d]
• The common difference (d) of an arithmetic progression can be positive, negative, or zero.
  • Positive d: The sequence is increasing, for example: 2, 5, 8, 11, …
    
  • Negative d: The sequence is decreasing, for example: 8, 5, 2, −1, …
    
  • Zero d: The sequence remains constant, for example: 4, 4, 4, 4, …
    
• The graphical representation of an arithmetic progression is a straight line, and the slope of this line is equal to the common difference (d).
• If an arithmetic progression has an even number of terms, the middle term is equal to the average of the first and last terms.
• This article includes chapter notes, formulas, exercise links, and important questions for Chapter 5 – Arithmetic Progressions.
• Class 10 Mathematics Chapter 5 contains four exercises, comprising a total of 49 fully solved questions.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Introduction

In this chapter, students will learn about special number patterns. They will understand how to identify these patterns, find specific terms in a given sequence, and calculate the sum of all numbers in a sequence.

Arithmetic Progression

In simple words, an Arithmetic Progression (AP) is a list of numbers arranged in a specific order, where the difference between any two consecutive numbers is the same.

Example:
The series 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is a sequence of natural numbers. This series is also an arithmetic progression because the difference between any two consecutive terms is 1. Similarly, in a series of odd numbers or even numbers, the difference between consecutive terms is 2. Hence, both odd and even number series are also arithmetic progressions.

There are three types of progressions discussed in Class 10 Maths Chapter 5:

• Arithmetic Progression (AP)
• Geometric Progression (GP)
• Harmonic Progression (HP)

Students should also remember the following definitions of Arithmetic Progression:

• Definition 1:
  An arithmetic progression (AP) is a sequence of numbers in which the difference between any two consecutive terms remains constant.
• Definition 2:
  In an arithmetic progression, each new term is obtained by adding a fixed number to the previous term.
• Definition 3:
  The common difference of an AP is the fixed number added to each term.
  For example, in the AP 1, 4, 7, 10, 13, 16, 19, 22, the common difference is 3.

In Class 10 Maths Chapter 5 Solutions, there are three main terms. These terms are:

• The common difference (d)
• nth Term (an)
• The num of the first n terms (Sn)
• These three terms are used to represent the property of arithmetic progression. In the next section, we will look at these three properties in more detail.

The Common Difference in Arithmetic Progression

For any arithmetic progression, the key elements used are the first term, the common difference between consecutive terms, and the nth term. Suppose a1, a2, a3, a4, …, an​ form an arithmetic progression.

This means that the value of the common difference ‘d’ is:

D = a2 − a1 = a3 − a2 = … = an − an−1

Here, d represents the common difference. Its value can be positive, negative, or zero.

The First Term of an Arithmetic Progression

If one wishes to express an arithmetic progression in terms of its common difference for solving purposes, it can be written as:

A, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …, a + (n – 1) d

In this sequence, a represents the first term of the arithmetic progression.

The general representation of an arithmetic progression

In this section, students will learn how to do exactly that. Before proceeding, students should start by assuming that the arithmetic progression for Class 10 Maths Chapter 5 is: a1, a2, a3, …, an  

Position of Terms	Representation of Terms	Values of Terms
1	a1	A = a + (1 − 1) d
2	a2	A + d = a + (2 − 1) d
3	a3	A + 2d = a + (3 − 1) d
4	a4	A + 3d = a + (4 − 1) d
…	…	…
n	an	A + (n − 1) d

The nth Term of Arithmetic Progression (AP)

This formula is used to find the Class 10 Maths Chapter 5 NCERT solutions where the value of the nth term of an arithmetic progression is required. The formula can be expressed as:

An = a + (n − 1)d

Here, a is the first term, d is the value of the common difference, n is the number of terms, and an is the nth term.

Let us consider an example. Find the nth term of the arithmetic progression
1, 2, 3, 4, 5, …, a_n, where the total number of terms is 15.

We know that n = 15. Therefore, using the formula, we can say that:

An = a + (n − 1)d

Since, a = 1, and the common difference or d = 2 – 1 = 1

Then, an = 1 + (15 − 1)1 = 1 + 14 = 15.

The whole series’ characteristics are totally dependent on the common difference’s value. When the common difference is positive, the sequence terms grow larger and push towards positive infinity. On the other hand, if the common difference is negative, the sequence terms get smaller and push towards negative infinity.

The Sum of the First n Terms of an Arithmetic Progression(AP)

One can easily calculate the sum of n terms of a given progression. In the case of an arithmetic progression, the sum of the first n terms can be found when the first term and the number of terms are known. The formula for this is given below.

S = n/2 (2a + (n − 1) × d)

But what if the value of the last term of the arithmetic progression is given? In that case, students should use the formula that is mentioned below.

S = n/2 (first term + last term)

For ease of revision, we have also summarized all the major formulas of this chapter in a table. That table is mentioned below.

General Form of AP	a, a + d, a + 2d, a + 3d, a + 4d, … , a + nd
The nth term of AP	An = a + (n − 1) × d
Sum of n terms in AP	S = n⁄2 (2a + (n − 1) × d)
Sum of all terms in a finite AP with the last term as l	S = N⁄2 (a + l)

Conclusion

NCERT Solutions for Class 10 Maths Chapter on Arithmetic Progressions by ToppersSky help students build a strong and clear understanding of this essential topic. Through interactive and animated learning, students can easily grasp key concepts such as the nth term, sum of terms, and common difference. The step-by-step explanations combined with visual animations make complex ideas simple and engaging, helping students understand the logic behind each formula and method. This approach not only improves concept clarity but also strengthens problem-solving skills.

Arithmetic Progressions form the foundation for many advanced mathematical topics, so mastering this chapter is crucial. In previous board exams, around five to six questions are typically asked from this chapter. By practising NCERT-based problems along with previous years’ questions on ToppersSky, students can boost their confidence, improve accuracy, and prepare effectively for exams.

Class 10 Maths Chapter 5: Arithmetic Progressions: Practice & Exercises

Q. 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.

Ans: Given the fare of first km is Rs. 15 and the fare for each additional km is Rs. 8. Hence,

Taxi fare for 1st km is Rs. 15.

Taxi fare for 2nd km is Rs. 15 + 8 = 23.

Taxi fare for 3rd km is Rs. 23 + 8 = 31.

Similarly, Taxi fare for nth km is Rs. 15 + (n − 1)8.

Therefore, we can conclude that the above list forms an A.P with common difference of 8.

(ii). The Amount of Air Present in a Cylinder When a Vacuum Pump Removes a Quarter of the Air Remaining in the Cylinder at a Time.

Ans: Let the initial volume of air in a cylinder be V liter. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. Hence,

Volume after 1st stroke is 3V/4.
> Volume after 2nd stroke is (3/4)(3V/4).
> Volume after 3rd stroke is (3/4)² (3V/4).

Similarly, volume after nth stroke is (3/4)ⁿ V.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by 3/4. Therefore, we can conclude that the above list does not form an A.P.

(iii). The cost of digging a well after every meter of digging, when it costs Rs. 150 for the first meter and rises by Rs. 50 for each subsequent meter.

Ans: Given the cost of digging for the first meter is Rs. 150 and the cost for each additional meter is Rs. 50. Hence,

Cost of digging for 1st meter is Rs. 150.
Cost of digging for 2nd meter is Rs. 150 + 50 = 200.
Cost of digging for 3rd meter is Rs. 200 + 50 = 250.

Similarly, cost of digging for nth meter is Rs. 150 + (n − 1)50.

Therefore, we can conclude that the above list forms an A.P with common difference of 50.

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

Ans: Given the principal amount is Rs. 10000 and the compound interest is 8% per annum. Hence,

Amount after 1st year is Rs. 10000 (1 + 8/100).

Amount after 2nd year is Rs. 10000 (1 + 8/100)2.

Amount after 3rd year is Rs. 10000 (1 + 8/100)3.

Similarly, amount after nth year is Rs. 10000 (1 + 8/100)n.

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by (1 + 8/100). Therefore, we can conclude that the above list does not form an A.P.

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

1. a = 10, d = 10

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(1)
Substituting a = 10, d = 10 in (1), we get,
an = 10 + 10(n − 1) = 10n   …(2)
Therefore, from (2)
a1 = 10, a2 = 20, a3 = 30 and a4 = 40.

2. a = −2, d = 0

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(1)
Substituting a = −2, d = 0 in (1), we get,
an = −2 + 0(n − 1) = −2   …(2)
Therefore, from (2)
a1 = −2, a2 = −2, a3 = −2 and a4 = −2.

3. a = 4, d = −3

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(1)
Substituting a = 4, d = −3 in (1), we get,
an = 4 − 3(n − 1) = 7 − 3n   …(2)
Therefore, from (2)
a1 = 4, a2 = 1, a3 = −2 and a4 = −5.

4. a = −1, d = 1/2

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(1)
Substituting a = −1, d = 1/2 in (1), we get,
an = −1 + 1/2(n − 1) = (n − 3)/2   …(2)
Therefore, from (2)
a1 = −1, a2 = −1/2, a3 = 0 and a4 = 1/2.

5. a = −1.25, d = −0.25

Ans: We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(1)
Substituting a = −1.25, d = −0.25 in (1) we get,
an = −1.25 − 0.25(n − 1) = −1 − 0.25n   …(2)
Therefore, from (2)
a1 = −1.25, a2 = −1.5, a3 = −1.75 and a4 = −2.

Q. 3. For the following A.P.s, write the first term and the common difference.

1. 3, 1, −1, −3, …

Ans: From the given A.P., we can see that the first term is 3.
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference = 2nd term − 1st term
∴ Common difference = 1 − 3 = −2.

2. −5, −1, 3, 7, …

Ans: From the given A.P., we can see that the first term is −5.
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference = 2nd term − 1st term
∴ Common difference = −1 − (−5) = 4.

3. 1/3, 5/3, 9/3, 13/3, …

Ans: From the given A.P., we can see that the first term is 1/3.
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference = 2nd term − 1st term
∴ Common difference = 5/3 − 1/3 = 4/3.

4. 0.6, 1.7, 2.8, 3.9, …

Ans: From the given A.P., we can see that the first term is 0.6.
The common difference is the difference between any two consecutive numbers of the A.P.
Common difference = 2nd term − 1st term
∴ Common difference = 1.7 − 0.6 = 1.1.

Q. 4. Which of the following are AP’s? If they form an AP, find the common difference d and write three more terms.

1. 2, 4, 8, 16, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 4 − 2 = 2    …(1)
a3 − a2 = 8 − 4 = 4    …(2)
a4 − a3 = 16 − 8 = 8    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.

2. 2, 5⁄2, 3, 7⁄2, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 5⁄2 − 2 = 1⁄2 …(1)
a3 − a2 = 3 − 5⁄2 = 1⁄2 …(2)
a4 − a3 = 7⁄2 − 3 = 1⁄2 …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term 2 and common difference 1⁄2.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = 2, d = 1⁄2 in (4), we get:
an = 2 + 1⁄2(n − 1) = n + 3⁄2 …(5)
Therefore, from (5):
a5 = 4, a6 = 9⁄2, and a7 = 5.

3. 1.2, 3.2, 5.2, 7.2, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 3.2 − 1.2 = 2    …(1)
a3 − a2 = 5.2 − 3.2 = 2    …(2)
a4 − a3 = 7.2 − 5.2 = 2    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term 1.2 and common difference 2.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = 1.2, d = 2 in (4), we get:
an = 1.2 + 2(n − 1) = 2n − 0.8    …(5)
Therefore, from (5):
a5 = 9.2, a6 = 11.2 and a7 = 13.2.

4. −10, −6, −2, 2, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = −6 − (−10) = 4    …(1)
a3 − a2 = −2 − (−6) = 4    …(2)
a4 − a3 = 2 − (−2) = 4    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term −10 and common difference 4.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = −10, d = 4 in (4), we get:
an = −10 + 4(n − 1) = 4n − 14    …(5)
Therefore, from (5):
a5 = 6, a6 = 10 and a7 = 14.

5. 3, 3 + √2, 3 + 2√2, 3 + 3√2, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = (3 + √2) − 3 = √2    …(1)
a3 − a2 = (3 + 2√2) − (3 + √2) = √2    …(2)
a4 − a3 = (3 + 3√2) − (3 + 2√2) = √2    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term 3 and common difference √2.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = 3, d = √2 in (4), we get:
an = 3 + (n − 1)√2    …(5)
Therefore, from (5):
a5 = 3 + 4√2, a6 = 3 + 5√2 and a7 = 3 + 6√2.

6. 0.2, 0.22, 0.222, 0.2222, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 0.22 − 0.2 = 0.02    …(1)
a3 − a2 = 0.222 − 0.22 = 0.002    …(2)
a4 − a3 = 0.2222 − 0.222 = 0.0002    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.

7. 0, −4, −8, −12, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = −4 − 0 = −4    …(1)
a3 − a2 = −8 − (−4) = −4    …(2)
a4 − a3 = −12 − (−8) = −4    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term 0 and common difference −4.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = 0, d = −4 in (4), we get:
an = 0 − 4(n − 1) = 4 − 4n    …(5)
Therefore, from (5):
a5 = −16, a6 = −20 and a7 = −24.

8. −1⁄2, −1⁄2, −1⁄2, −1⁄2, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = (−1⁄2) − (−1⁄2) = 0    …(1)
a3 − a2 = (−1⁄2) − (−1⁄2) = 0    …(2)
a4 − a3 = (−1⁄2) − (−1⁄2) = 0    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term −1⁄2 and common difference 0.
We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = −1⁄2, d = 0 in (4), we get:
an = −1⁄2 + 0(n − 1) = −1⁄2 …(5)
Therefore, from (5):
a5 = −1⁄2, a6 = −1⁄2 and a7 = −1⁄2.

9. 1, 3, 9, 27, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 3 − 1 = 2    …(1)
a3 − a2 = 9 − 3 = 6    …(2)
a4 − a3 = 27 − 9 = 18    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.

10. a, 2a, 3a, 4a, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = 2a − a = a    …(1)
a3 − a2 = 3a − 2a = a    …(2)
a4 − a3 = 4a − 3a = a    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term a and common difference a.

We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = a, d = a in (4), we get:
an = a + (n − 1)a    …(5)
Therefore:
a5 = a + (5 − 1)a = 5a
a6 = a + (6 − 1)a = 6a
a7 = a + (7 − 1)a = 7a

11. a, a2, a3, a4, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = a2 − a = a(a − 1)    …(1)
a3 − a2 = a3 − a2 = a2(a − 1)    …(2)
a4 − a3 = a4 − a3 = a3(a − 1)    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.
Therefore, the given series does not form an A.P.

12. √2, √8, √18, √32, …

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.
a2 − a1 = √8 − √2 = 2√2 − √2 = √2    …(1)
a3 − a2 = √18 − √8 = 3√2 − 2√2 = √2    …(2)
a4 − a3 = √32 − √18 = 4√2 − 3√2 = √2    …(3)
From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.
Therefore, the given series forms an A.P. with first term √2 and common difference √2.

We know that the nth term of the A.P. with first term a and common difference d is given by:
an = a + (n − 1)d    …(4)
Substituting a = √2, d = √2 in (4), we get:
an = √2 + (n − 1)√2
Therefore:
a5 = √2 + 4√2 = 5√2 = √50
a6 = √2 + 5√2 = 6√2 = √72
a7 = √2 + 6√2 = 7√2 = √98

Q.5. Choose the correct choice in the following and justify

(i). 30th term of the A.P. 10, 7, 4, …, is
1. 97
2. 77
3. −77
4. 87

Ans: C. −77
Given, the first term, a = 10 …(1)
Given, the common difference, d = 7 − 10 = −3 …(2)
Given, the number of terms, n = 30 …(3)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d …(4)
Substituting the values from (1), (2) and (3) in (4) we get,
an = 10 + (30 − 1)(−3)
⇒ an = 10 − 87
∴ an = −77

(ii). 11th term of the A.P. −3, −1⁄2, 2, …, is
1. 28
2. 22
3. 38
4. 481⁄2

Ans: B. 22
Given, the first term, a = −3 …(1)
Given, the common difference, d = −1⁄2 − (−3) = 5⁄2 …(2)
Given, the number of terms, n = 11 …(3)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d …(4)
Substituting the values from (1), (2) and (3) in (4) we get,
an = −3 + 5⁄2(11 − 1)
⇒ an = −3 + 25
∴ an = 22

Q.6. Which term of the A.P. 3, 8, 13, 18, … is 78?

Ans: Given, the first term a = 3 …(1)
Given, the common difference d = 8 − 3 = 5 …(2)
Given, the nth term, an = 78 …(3)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d …(4)
Substituting the values from (1), (2) and (3) in (4) we get,
78 = 3 + 5(n − 1)
⇒ 75 = 5(n − 1)
⇒ 15 = (n − 1)
∴ n = 16
Therefore, the 16th term of this A.P. is 78.

Q. 7. Check whether −150 is a term of the A.P. 11, 8, 5, 2, …

Ans: Given, the first term a = 11 …(1)
Given, the common difference d = 8 − 11 = −3 …(2)
Given, the nth term an = −150 …(3)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d …(4)
Substituting the values from (1), (2) and (3) in (4) we get,
−150 = 11 − 3(n − 1)
⇒ −161 = −3(n − 1)
⇒ 161⁄3 = (n − 1)
∴ n = 164⁄3
Since n is not a natural number, therefore −150 is not a term of the given A.P. series.

Q. 8. If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Ans: Given that the 17th term of an A.P. exceeds its 10th term by 7 i.e.,
a17 = a10 + 7   …(1)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d   …(2)
For 17th term substitute n = 17 in (2) i.e.,
a17 = a + 16d   …(3)
For 10th term substitute n = 10 in (2) i.e.,
a10 = a + 9d   …(4)
Therefore, from (1), (3) and (4) we get,
a + 16d = a + 9d + 7
⇒ 7d = 7
∴ d = 1
Therefore, the common difference is 1.

Q. 9.Determine the A.P. whose third term is 16 and the 7ᵗʰ term exceeds the 5ᵗʰ term by 12.

Ans: Given the 7ᵗʰ term of the A.P. is 12 more than its 5ᵗʰ term, i.e.,
a₇ = a₅ + 12 …(1)
We know that the nᵗʰ term of the A.P. with first term a and common difference d is given by:
aₙ = a + (n − 1)d …(2)
For the 5ᵗʰ term, substitute n = 5 in (2):
a₅ = a + 4d …(3)
For the 7ᵗʰ term, substitute n = 7 in (2):
a₇ = a + 6d …(4)

Therefore, from (1), (3), and (4), we get:
a + 6d = a + 4d + 12
⇒ 2d = 12
∴ d = 6 …(5)
Substituting (5) in (2), we get:
aₙ = a + 6(n − 1) …(6)
Given the third term of the A.P. is 16. Hence from (6):
16 = a + 6(3 − 1)
⇒ 16 = a + 12
∴ a = 4 …(7)
Hence from (6),
aₙ = 4 + 6(n − 1)
Therefore, the A.P. is:
4, 10, 16, 22, …

Q. 10. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Ans: Given, first term, a = 17  ….(1)
Given, the common difference, d = 9  ….(2)
Given, the nth term, an = 350  ….(3)
We know that the nth term of the A.P. with first term a and common difference d is given by
an = a + (n − 1)d  ….(4)
Substituting the values from (1), (2), (3) in (4) we get,
350 = 17 + 9(n − 1)
⇒ 333 = 9(n − 1)
⇒ 37 = (n − 1)
∴ n = 38  ….(5)
We know that the sum of n terms of the A.P. with first term a and last term l is given by
Sn = n/2 [a + l]  ….(6)
Substituting the values from (1), (5), (3) in (6) we get,
S38 = 38/2 [17 + 350]
⇒ S38 = 19 (367)
∴ S38 = 6973

FAQs

1. Must the common difference ‘d’ in an AP always be a positive number?

No, the common difference (d) in an arithmetic progression does not have to be positive. It can be positive, negative, or zero. The only condition is that the difference between any two consecutive terms must remain constant throughout the sequence.

2. Do the NCERT Solutions for Arithmetic Progressions Class 10 only give the final answers?

The NCERT Solutions for Class 10 Maths Chapter 5 provide complete solution explanations which cover all mathematical problems in the chapter through detailed step-by-step solutions. The solution system displays all aspects of problem-solving including the required formulas and their respective computational processes.

3. If I use NCERT solutions, won’t I become dependent on them and stop learning?

The smart study method of using Class 10 Maths Chapter 5 question answer solutions functions as an effective learning tool because it supports educational activities. The key to problem-solving lies in using answer keys which should not be copied for direct solution purposes but should serve as tools to correct and explain your work after you have attempted to resolve the issues independently.

4. Is the ‘nth term’ formula all I need for this chapter?

The ‘nth term’ formula shows you the main concept of arithmetic sequences but you need ‘sum of n terms’ formulas because they help you solve arithmetic progressions questions and answers in class 10. The formulas needed for this purpose are S_n = n/2[2a + (n-1)d] and S_n = n/2[a + l].

5. Are word problems in Arithmetic Progressions too complicated and not based on formulas?

The Class 10 Maths Chapter 5 word problems about Arithmetic Progressions require students to solve them with standard AP formulas according to their design. The main challenge of the problem exists because people must transform its information into essential parts of an AP sequence: which include its first term (a), common difference (d), total terms (n), specific term (a_n), and sum (S_n) components.