NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions : Exercise 5.1

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.1 holds the key to developing a robust concept of Arithmetic Progressions (AP) and the related ideas. Knowing this exercise is a foundation for the more difficult mathematical concepts in the higher classes. In this analysis, we're going to discuss the most critical aspects of AP and walk you through these essential concepts while aligning them with the current CBSE syllabus. Let's delve into this fundamental principle that underlies numerous practical uses and advanced math studies.

1.0NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 : Free PDF

2.0Introduction to Arithmetic Progression

In the world outside, most patterns occur in a certain sequence. For example, the petals on a sunflower, the spirals on a pinecone, or even the increase in salary over time can all be described with arithmetic progressions. Arithmetic Progression (AP) refers to a collection of numbers that are known as the terms of an AP. Every element of this AP is determined through the addition of a specific value, denoted as the common difference (d), to its previous element. AP can either be infinite or finite.

Common Difference

Common Difference in an Arithmetic Progression (AP) is the constant number that is added (or deducted) to a term to obtain the succeeding term. It does not change within the sequence. The common difference is represented by d. Note that common differences can be positive, negative, or even zero in an AP. 

Types of Arithmetic Progression

1. Definite Arithmetic Progression (AP): Definite Arithmetic Progression (AP) is an AP that has a limited number of terms. In simple words, it has a last term, and the sequence of the AP ends after a finite number of terms, for example, 5, 10, 15, 20, and 25. In this sequence, 25 is the last term and has a finite number of terms, which is 5. 
2. Indefinite Arithmetic Progression (AP): An infinite arithmetic progression is an AP with an infinite number of terms. In other words, the sequence goes on forever, so it has no last term and extends indefinitely in both directions (negative or positive). For instance, 2, 4, 6, 8, 10, …. In this example, the sequence goes on forever, and we don’t know the last term of this AP. 

3.0Exercise 5.1 Overview: Key Concepts 

How to Find Common Differences

To find the value of the common difference, we subtract any term of the AP from the term that follows it. The common difference is denoted by the letter d and can be found by using the formula: 

$d=a_{n+1}-a_n$

Here, 

• a_n+1 is the (n+1)th term (the next term),
• a_n is the nth term (the current term).

Identifying an AP from a List of Numbers

To identify an AP from a list of numbers, follow the steps mentioned below:

• Verify Differences Between Successive Terms: Deduce the differences between consecutive terms by subtracting the first term from the second term, the second term from the third term, and so on.
• Check whether the Differences are Constant: If the difference between each consecutive term remains constant, then the sequence of numbers is an AP. This common difference is referred to as such. 

Formation of an Arithmetic Progression

The formation of an AP is an important concept of the first exercise of chapter 5. This can be simply done by using these steps: 

• Decide the first term of your AP. This is the number, from where the AP begins. 
• Determine the common difference of an AP if not given beforehand. 
• Add this common difference to the first term to get the second term. 
• Continue till you get the required number of terms in the AP. 

4.0NCERT Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 : Detailed Solutions

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 ₹ 15 for the first km and ₹ 8 for each additional km.

(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.

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

(iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.

Solution

(i) Let tₙ denote the taxi fare (in ₹) for the first n km.

Now, t₁ = 15,

t₂ = 15 + 8 = 23,

t₃ = 23 + 8 = 31,

t₄ = 31 + 8 = 39, ...

List of fares after 1 km, 2 km, 3 km, 4 km, ... respectively is 15, 23, 31, 39, .... (in ₹).

Here, t₂ -t₁ = t₃ - t₂ =  t₄ - t₃ = ... = 8. Thus, the list forms an AP.

(ii) Let t₁ = x units; t₂ = x - (1/4)x = (3/4)x units;

t₃ = (3/4)x - (1/4)(3/4)x = (9/16)x units;

t₄ = (9/16)x - (1/4)(9/16)x = (27/64)x units; ...

The list of numbers is x, (3/4)x, (9/16)x, (27/64)x, ....

It is not an AP because t₂ - t₁ ≠t₃ - t₂.

(iii) Cost of digging for first metre = ₹ 150

Cost of digging for first 2 metres = 150 + 50 = ₹ 200

Cost of digging for first 3 metres = 200 + 50 = ₹ 250

Cost of digging for first 4 metres = 250 + 50 = ₹ 300

Clearly, 150, 200, 250, 300.... forms an AP.

Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = ... . Thus, the list forms an AP.

(iv) We know that if ₹ P is deposited at r% compound interest per annum for n years, our money will be P(1 + r/100) after n years.

Therefore, after every year, our money will be

10000(1 + 8/100), 10000(1 + 8/100)²,

10000(1 + 8/100)³, 10000(1 + 8/100)⁴, ...

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an AP.

2. Write first four terms of the AP, when the first term a and the common difference d 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

Solution

(i) t₁ = a = 10,

t₂ = 10 + d = 10 + 10 = 20,

t₃ = 20 + d = 20 + 10 = 30,

t₄ = 30 + d = 30 + 10 = 40

Thus, first four terms of the AP are 10, 20, 30, 40

(ii) Given a = -2 and d = 0

t₁ = -2, t₂ = -2 + 0 = -2,

t₃ = -2 + 0 = -2, t₄ = -2 + 0 = -2

Thus, first four terms of the AP are -2, -2, -2, -2

(iii) a = 4, d = -3

t₁ = a = 4

t₂ = t₁ + d = 4 - 3 = 1

t₃ = t₂ + d = 1 - 3 = -2

t₄ = t₃ + d = -2 - 3 = -5

Thus, the first four terms of the AP are 4, 1, -2 and -5.

(iv) a = -1, d = 1/2

t₁ = a = -1

t₂ = t₁ + d = -1 + 1/2 = -1/2

t₃ = t₂ + d = -1/2 + 1/2 = 0

t₄ = t₃ + d = 0 + 1/2 = 1/2

Thus, the first four terms of the AP are -1, -1/2, 0 and 1/2.

(v) a = -1.25, d = -0.25

t₁ = a = -1.25

t₂ = t₁ + d = -1.25 - 0.25 = -1.50

t₃ = t₂ + d = -1.50 - 0.25 = -1.75

t₄ = t₃ + d = -1.75 - 0.25 = -2.00

Thus, the first four terms of the AP are -1.25, -1.50, -1.75 and -2.00.

3. For the following APs, 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, ...

Solution

(i) a = 3, d = t₂ - t₁ = 1 - 3 = -2, i.e., d = -2

(ii) a = -5, d = t₂ - t₁ = -1 - (-5) = 4

(iii) a = 1/3, d = t₂ - t₁ = 5/3 - 1/3 = 4/3

(iv) a = 0.6, d = t₂ - t₁ = 1.7 - 0.6 = 1.1

4. Which of the following are APs? If they form an AP, 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, ...

Sol.

(i) Not an AP because t₂ - t₁ = 2 and t₃ - t₂ = 8 - 4 = 4, i.e., t₂ - t₁ ≠ t₃ - t₂.

(ii) It is an AP.

a = 2, d = 1/2

[∵ t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = 1/2]

t₅ = 7/2 + 1/2 = 4, t₆ = 4 + 1/2 = 9/2, tℏ = 9/2 + 1/2 = 5

(iii) We have -1.2, -3.2, -5.2, -7.2, ...

∴ t₁ = -1.2, t₂ = -3.2, t₃ = -5.2, t₄ = -7.2

t₂ - t₁ = -3.2 + 1.2 = -2

t₃ - t₂ = -5.2 + 3.2 = -2

t₄ - t₃ = -7.2 + 5.2 = -2

∵ t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = -2

⇒ d = -2

∴ The given numbers form an AP such that d = -2.

Now, t₅ = t₄ + (-2) = -7.2 + (-2) = -9.2,

t₆ = t₅ + (-2) = -9.2 + (-2) = -11.2 and

tℏ = t₆ + (-2) = -11.2 + (-2) = -13.2

Thus, d = -2 and t₅ = -9.2, t₆ = -11.2 and t₇ = -13.2.

(iv) It is an AP.

a = -10, d = 4, t₅ = 6, t₆ = 10, t₇ = 14.

(v) It is an AP.

a = 3, d = √2

t₅ = 3 + 3√2 + √2 = 3 + 4√2,

t₆ = 3 + 5√2, tℏ = 3 + 6√2.

(vi) It is not an AP.

t₂ - t₁ = 0.22 - 0.2 = 0.02,

t₃ - t₂ = 0.222 - 0.22 = 0.002, ...

i.e., t₂ - t₁ ≠ t₃ - t₂.

(vii) We have: 0, -4, -8, -12, ...

∴ t₁ = 0, t₂ = -4, t₃ = -8, t₄ = -12

t₂ - t₁ = -4 - 0 = -4

t₃ - t₂ = -8 + 4 = -4

t₄ - t₃ = -12 + 8 = -4

∵ t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = -4 ⇒ d = -4

∴ The given numbers form an AP.

Now, t₅ = t₄ + (-4) = -12 + (-4) = -16

t₆ = t₅ + (-4) = -16 + (-4) = -20

tℏ = t₆ + (-4) = -20 + (-4) = -24

Thus, d = -4 and t₅ = -16, t₆ = -20, t₇ = -24.

(viii) We have: -1/2, -1/2, -1/2, -1/2, ...

∴ t₁ = t₂ = t₃ = t₄ = -1/2

t₂ - t₁ = 0, t₃ - t₂ = 0, t₄ - t₃ = 0

⇒ d = 0

∴ The given numbers form an AP.

Now, t₅ = -1/2 + 0 = -1/2

t₆ = -1/2 + 0 = -1/2, tℏ = -1/2 + 0 = -1/2

Thus, d = 0 and t₅ = -1/2, t₆ = -1/2, tℏ= -1/2.

(ix) Not an AP. Here, t₂ - t₁ ≠ t₃ - t₂.

(x) We have: a, 2a, 3a, 4a, ...

∴ t₁ = a, t₂ = 2a, t₃ = 3a, t₄ = 4a

t₂ - t₁ = 2a - a = a,

t₃ - t₂ = 3a - 2a = a and

t₄ - t₃ = 4a - 3a = a

∵ t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = a

⇒ d = a

∴ The given numbers form an AP.

Now, t₅ = t₄ + a = 4a + a = 5a,

t₆ = t₅ + a = 5a + a = 6a

and tℏ = t₆ + a = 6a + a = 7a

Thus, d = a and t₅ = 5a, t₆ = 6a, t₇ = 7a.

(xi) Not an AP if a ≠ 1.

Here, t₂ - t₁ = a² - a = a(1 - a),

t₃ - t₂ = a³ - a² = a²(1 - a)

t₃ - t₂ ≠ t₂ - t₁ when a ≠ 1.

It will be an AP if a = 1.

Hence, the given sequence is an AP only when a = 1.

In this case, first term = 1, common difference = 0.

(xii) It is an AP.

√2, √8, √18, √32, ... can be rewritten as

√2, 2√2, 3√2, 4√2, ...

a = √2, d = √2

t₅ = 5√2, t₆ = 6√2, t<0xE2><0x82><0x87> = 7√2,

i.e., t₅ = √50, t₆ = √72, t₇= √98.

5.0Benefits of Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1

• Helps you recognize patterns in numbers.
• Enhances problem-solving skills with step-by-step calculations.
• Useful in finance, physics, and daily life scenarios like savings plans.
• Strengthens your ability to work with algebraic expressions.
• Forms the foundation for sequences and series in advanced studies.