NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions

NCERT Solutions for Chapter 5 of Class 10 Maths, Arithmetic Progressions, introduces students to the interesting world of sequences and patterns that can be found in everyday objects like pine cone spirals and flower petals. This chapter explores how each term in a sequence can be derived by adding a fixed number, known as the common difference, to the previous term. Understanding these concepts is crucial for solving real-life problems. 

The solutions provide a comprehensive overview of key terminologies, including 'term' and 'common difference', along with detailed explanations of how to find the 'n-th' term and the sum of 'n' consecutive terms. NCERT Class 10 Maths - Chapter 5 provides students with well-illustrated examples and formulas to help them better understand arithmetic progressions. This will help them cope with exercises and perform better on exams.

1.0Download NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions : Free PDF

The NCERT Solutions for Class 10 Maths Chapter 5 is available in the downloadable PDF format below:

2.0NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: Overview

• In Chapter 5 of Class 10 Maths, titled Arithmetic Progressions, students can find comprehensive solutions to help them grasp the concept of sequences and patterns that are commonly encountered in everyday life. 
• This chapter introduces the concept of an arithmetic progression (AP), a sequence of numbers in which the difference between consecutive terms remains consistent. The next terms in the sequence must be determined using this constant difference.
• Through detailed explanations and examples, students can clear up any confusion and improve their problem-solving abilities. 
• The chapter also includes exercises designed to reinforce learning, making it a valuable resource for mastering arithmetic progressions and effectively preparing for exams.

Subtopics of Chapter 5 Arithmetic Progression in Class 10

Below we have provided the list of topics that we will cover in the NCERT Solution Class 10 Chapter 5:

• Arithmetic Progressions
• nth Term of an AP
• The sum of the First n Terms of an AP 

3.0Chapter 5 Arithmetic Progressions NCERT Solutions for Class 10 Maths: All Exercises

The NCERT solutions offer clear answers to all exercises, helping students solve each problem confidently. By learning about concepts like common differences, the n-th term, and the sum of terms, students can prepare well for exams and gain a better understanding of how arithmetic progressions are used in math and real-life situations. Check the number of questions included in each exercise.

4.0Why You Choose NCERT Math Class 10 Chapter 5 Solutions?

There are a few reasons that will help you understand why you need to choose the NCERT Solution for Class 10 Math:

• NCERT Solution Class 10 Maths is the best material for students who want to understand the concepts of arithmetic progressions. 
• These solutions provide a clear and structured approach to understanding the various types of patterns and sequences that are fundamental in mathematics.
• NCERT Solutions are designed to cover every exercise so that students can practice extensively and build confidence in their problem-solving abilities.
• The solutions are aligned with the CBSE curriculum, making them an essential resource for exam preparation.
• NCERT Solutions are very effective, as they allow you to learn at your own pace, revisit challenging concepts, and enhance your understanding of arithmetic progressions, ultimately leading to improved academic performance.

5.0Important Topics of Class 10 Math Chapter 5 Arithmetic Progression Solutions

Chapter 5, Arithmetic Progression (AP), deals with sequences where the difference between consecutive terms is constant, known as the common difference (d). It covers important concepts like finding the nth term of an AP and the sum of the first n terms. Students also learn to apply these concepts in real-life problems, such as calculating distances or total amounts. Mastering AP helps in solving various mathematical problems and understanding sequences in algebra.

1. Introduction to Arithmetic Progression (AP)
2. General Form of an AP
3. Sum of First n Terms (Sₙ)
4. Applications of AP
5. Properties of AP

6.0Outline for Class 10 Maths: Arithmetic Progressions

Introduction to Arithmetic Progression (AP)

Definition: An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference.

General Form of an AP: The nth term of an AP is given by: $a_n=a+(n-1)\cdot d$ where:

• a is the first term,
• d is the common difference,
• n is the number of terms.

Examples of AP:

• 2, 5, 8, 11, $\dots$ with a common difference d = 3.
• 10, 7, 4, 1, $\dots$ with a common difference d = –3.

Important Terms in an AP

• First Term (a): The initial term in the progression.
• Common Difference (d): The difference between any two consecutive terms.
• Nth Term: The formula to find the nth term is a_n = a + (n-1) \cdot d.
• Number of Terms (n): The total number of terms in the sequence.
• Sum of Terms: The sum of the first n terms is denoted by S_n.

Derivation and Formula for the nth Term

• Formula for nth term: $a_n=a+(n-1)\cdot d$
• This formula helps find any term in an arithmetic progression when the first term, common difference, and the number of terms are known.

Sum of First n Terms of an AP

Formula for the sum of the first n terms $(S_n)$ of an arithmetic progression:

$S_n=\frac{n}{2}\cdot \left[2a+(n-1)\cdot d\right]$ or alternatively:

$S_n=\frac{n}{2}\cdot \left(a+l\right)$

where l is the last term of the AP.

Application: This formula is used to calculate the sum of a series of numbers in an arithmetic progression.

Properties of Arithmetic Progressions

• The Common Difference is constant between any two consecutive terms.
• Increasing AP: If d > 0, the sequence increases.
• Decreasing AP: If d < 0, the sequence decreases.
• Constant AP: If d = 0, all terms are the same.

Solving Problems Involving Arithmetic Progressions

• Finding the nth term of an AP.
• Determining the sum of the first n terms.
• Word problems involving real-life applications, such as calculating savings, distances, or time intervals.
• Examples:
• • Calculate the sum of the first 20 terms of the AP 3, 7, 11, 15, $\dots .$
  • Find the 15th term of the AP 5, 9, 13, 17, $\dots .$

7.0Sample NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

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. Fill in the blanks in the following table, given that a is the first term, d the common difference and $a_n$, the nth term of the AP.

Solution:

(i) a = 7, d = 3, n = 8

a₈ = a + 7d = 7 + 7 × 3 = 28.

Hence, a₈ = 28.

(ii) a = -18, n = 10, aₙ = 0, d = ?

aₙ = a + (n - 1)d

0 = -18 + (10 - 1)d

18 = 9d => d = 18/9 = 2

Hence, d = 2.

(iii) d = -3, n = 18, aₙ = -5

aₙ = a + (n - 1)d

-5 = a + (18 - 1)(-3)

-5 = a + (17)(-3)

-5 = a - 51

a = 51 - 5 = 46

Hence, a = 46.

(iv) a = -18.9, d = 2.5, aₙ = 3.6

=> a + (n - 1)d = 3.6

=> -18.9 + (n - 1) × (2.5) = 3.6

=> (n - 1) × (2.5) = 3.6 + 18.9 = 22.5

=> n - 1 = 22.5 / 2.5 = 225 / 25 = 9

=> n = 10

(v) a = 3.5, d = 0, n = 105

Then a₁₀₅ = a + 104d = 3.5 + 0 = 3.5

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. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Solution

$t_1$= ₹ 5 (savings in the 1st week)

$t_2$ = ₹ 5 + ₹ 1.75 = ₹ 6.75 (savings in the 2nd week)

$t_3$ = ₹ 6.75 + ₹ 1.75 = ₹ 8.50 (savings in the 3rd week)

$t_n$ = ₹ 20.75

=> a + (n - 1)d = 20.75

=> 5 + (n - 1) × 1.75 = 20.75

=> (n - 1) × 1.75 = 15.75

=> n - 1 = 15.75 / 1.75 = 1575 / 175 = 9

=> n = 10

Hence, in the 10th week, Ramkali's savings will be ₹ 20.75.

5. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Solution

It can be observed that the incomes that Subba Rao obtained in various years are in AP as every year, his salary is increased by ₹ 200.

Therefore, the salaries of each year after 1995 are

5000, 5200, 5400, ...

Here, a = 5000

d = 200

Let after nth year, his salary be ₹ 7000.

Therefore, an = a + (n - 1)d

7000 = 5000 + (n - 1)200

200(n - 1) = 2000

(n - 1) = 10

n = 11

Therefore, in the 11th year, his salary will be ₹ 7000.

8.0Benefits of Class 10 Maths Chapter 5 Arithmetic Progression 

Arithmetic Progression (AP) offers several benefits:

1. Simple Pattern: The terms follow a consistent, easy-to-understand pattern with a fixed difference between consecutive terms.
2. Financial Applications: AP is used to calculate regular payments, such as loan installments or savings plans.
3. Scheduling: Helps in planning events or tasks at regular intervals.
4. Real-Life Modeling: AP models situations like population growth or asset depreciation where changes occur at a constant rate.
5. Education: It’s foundational for understanding more complex sequences and series in mathematics.
6. Efficient Problem Solving: AP simplifies calculations in problems involving time, distance, or speed.
7. Resource Allocation: Useful in optimizing processes with evenly spaced intervals.

Overall, AP is easy to work with and has practical uses in finance, education, and real-world scenarios.