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NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions Exercise 5.4

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NCERT Solutions for Maths Class 10 Chapter 5 Exercise 5.4 - FREE PDF Download

Chapter 5 Arithmetic Progressions Exercise 5.4 Class 10, focuses on applying the concepts of arithmetic progressions (AP) to solve various problems. This exercise explains how to find the sum of the first n terms of an arithmetic progression. Students will learn how to use the formula for the sum of an AP and apply it to different types of problems. Students can access the revised Class 10 Maths NCERT Solutions from our page, which is prepared in such a way that you can understand it easily.

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Table of Content
1. NCERT Solutions for Maths Class 10 Chapter 5 Exercise 5.4 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 5 Exercise 5.4 Class 10 | Vedantu
3. Access NCERT Solutions for Maths Class 10 Chapter 5 - Arithmetic Progressions
    3.1Exercise 5.4
4. Class 10 Maths Chapter 5: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 5 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
7. NCERT Study Resources for Class 10 Maths
FAQs


These solutions are aligned with the updated CBSE guidelines for Class 10, ensuring students are well-prepared for exams. Class 10 Chapter 5 Maths Exercise 5.4 Questions and Answers PDF provides accurate answers to textbook questions and assists in effective exam preparation and better performance. Access the latest CBSE Class 10 Maths Syllabus here.


Glance on NCERT Solutions Maths Chapter 5 Exercise 5.4 Class 10 | Vedantu

  • In NCERT Solution for Class 10 Maths Chapter 5 Exercise 5.4 we will cover the topics related to the Arithmetic Progressions (AP).

  • Identify the first negative term: In an AP with decreasing terms, students will discover the term that dips below zero.

  • Students will learn to calculate the sum of terms at particular positions within the AP.

  • The exercise might involve scenarios like calculating the total length of wood needed for a ladder with rungs of decreasing size.

  • Students will encounter a challenging question about house numbers, where the sum of numbers preceding a specific house equals the sum of numbers following it.

  • Class 10 Exercise 5.4 contains 5 fully solved questions and solutions

Competitive Exams after 12th Science
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NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions Exercise 5.4
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Access NCERT Solutions for Maths Class 10 Chapter 5 - Arithmetic Progressions

Exercise 5.4

1. Which term of the A.P. \[\mathbf{121},\mathbf{117},\mathbf{113},...\] is its first negative term?

(Hint: Find $n$ for ${{a}_{n}}  <0$)

Ans: Given A.P. \[\mathbf{121},\mathbf{117},\mathbf{113},...\] 

Its first term is $121$ and the common difference is $117-121=-4$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$.

Therefore the ${{n}^{th}}$ term of the given A.P. is ${{a}_{n}}=121-4\left( n-1 \right)$   ….. (1)

To find negative term, find $n$ such that ${{a}_{n}} <0$  

Hence from (1),

$121-4\left( n-1 \right) <0$

$\Rightarrow 121 <4\left( n-1 \right)$ 

\[\Rightarrow \dfrac{121}{4}+1 < n\]

$\Rightarrow n >\dfrac{125}{4}$ 

$\therefore n >31.25$ 

Therefore, the ${{32}^{nd}}$ term of the given A.P. will be its first negative term.


2. The sum of the third and the seventh terms of an A.P is \[6\] and their product is \[8\]. Find the sum of first sixteen terms of the A.P.

Ans: Given the sum of third and seventh term of A.P., ${{a}_{3}}+{{a}_{7}}=6$  …..(1)

Given the sum of third and seventh term of A.P., ${{a}_{3}}\cdot {{a}_{7}}=8$  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$. Therefore, 

For $n=3,{{a}_{3}}=a+2d$ 

For $n=7,{{a}_{7}}=a+6d$ 

From (1), ${{a}_{3}}+{{a}_{7}}=\left( a+2d \right)+\left( a+6d \right)$ 

$\Rightarrow 2a+8d=6$ 

$\therefore a+4d=3$   ….. (3)

From (2), ${{a}_{3}}\cdot {{a}_{7}}=\left( a+2d \right)\cdot \left( a+6d \right)$ 

$\therefore {{a}^{2}}+8ad+12{{d}^{2}}=8$   …..(4)

Let us now solve equations (3) and (4) by substituting the value of $a$ from (3) into (4).

${{\left( 3-4d \right)}^{2}}+8d\left( 3-4d \right)+12{{d}^{2}}=8$

$\Rightarrow 9-24d+16{{d}^{2}}+24d-32{{d}^{2}}+12{{d}^{2}}=8$ 

$\Rightarrow -4{{d}^{2}}+1=0$

$\Rightarrow {{d}^{2}}=\dfrac{1}{4}$

$\therefore d=\dfrac{1}{2},-\dfrac{1}{2}$ …..(5)


CASE 1: 

For $d=\dfrac{1}{2}$ 

 Substitute $d=\dfrac{1}{2}$ in (6) we get, $a=1$ …..(6)

Therefore, it is an A.P series with first term $1$ and common difference $\dfrac{1}{2}$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, 

${{S}_{16}}=\dfrac{16}{2}\left[ 2+\dfrac{1}{2}\left( 16-1 \right) \right]$

$\Rightarrow {{S}_{16}}=4\left[ 19 \right]$ 

$\therefore {{S}_{16}}=76$ 


CASE 2: 

For $d=-\dfrac{1}{2}$ 

 Substitute $d=-\dfrac{1}{2}$ in (6) we get, $a=5$ …..(7)

Therefore, it is an A.P series with first term $5$ and common difference $-\dfrac{1}{2}$ and hence, 

${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2\left( 5 \right)-\dfrac{1}{2}\left( 16-1 \right) \right]$

$\Rightarrow {{S}_{16}}=4\left[ 5 \right]$ 

$\therefore {{S}_{16}}=20$ 


3. A ladder has rungs \[\mathbf{25}\] cm apart. (See figure). The rungs decrease uniformly in length from \[\mathbf{45}\] cm at the bottom to \[\mathbf{25}\] cm at the top. If the top and bottom rungs are $2\dfrac{1}{2}$ m apart, what is the length of the wood required for the rungs? 

(Hint: Number of Rungs $=\dfrac{250}{25}$)


Ladder with rungs 25cm apart


Ans: Distance between first and last rungs is $2\dfrac{1}{2}m=\dfrac{5}{2}m=250cm$.

Distance between two consecutive rungs is $25cm$.

Therefore, total number of rungs are $\dfrac{250}{25}+1=11$.

Also, we can observe that the length of each rung is decreasing in a uniform order. So, we can conclude that the length of rungs is in A.P. with first term $45$, common difference $-25$ and number of terms $11$.   

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$. Therefore, 

${{S}_{11}}=\dfrac{11}{2}\left[ 45+25 \right]$

$\Rightarrow {{S}_{11}}=11\left[ 35 \right]$ 

$\therefore {{S}_{11}}=385$

Therefore, the length of the wood required for the rungs is \[385\]cm.


4. The houses of a row are number consecutively from \[1\] to \[49\]. Show that there is a value of \[x\] such that the sum of numbers of the houses preceding the house numbered \[x\] is equal to the sum of the number of houses following it.

Find this value of \[x\].

(Hint \[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\])

Ans: Given houses are numbered $1,2,3,4,....$

Clearly, they are numbered in A.P. series with both first term and common difference as $1$.

Now, there is house numbered $x$ such that the sum of numbers of the houses preceding the house numbered \[x\] is equal to the sum of the number of houses following it i.e., \[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\]


Presentation of the house number on line


We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$. Therefore, 

\[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\]

$\Rightarrow \left\{ \dfrac{\left( x-1 \right)}{2}\left[ 1+\left( x-1 \right) \right] \right\}=\left\{ \dfrac{49}{2}\left[ 1+49 \right] \right\}-\left\{ \dfrac{x}{2}\left[ 1+x \right] \right\}$

$\Rightarrow \dfrac{x\left( x-1 \right)}{2}=49\left[ 25 \right]-\dfrac{x\left( x+1 \right)}{2}$ 

$\Rightarrow x\left( x-1 \right)=2450-x\left( x+1 \right)$

$\Rightarrow 2{{x}^{2}}=2450$

$\therefore x=35$   (Since house number cannot be negative)

Therefore, house number \[35\] is such that the sum of the numbers of houses preceding the house numbered \[35\] is equal to the sum of the numbers of the houses following it.


5. A small terrace at a football ground comprises of \[15\] steps each of which is \[50\] m long and built of solid concrete. Each step has a rise of $\dfrac{1}{4}$m and a tread of $\dfrac{1}{2}$m (See figure) calculate the total volume of concrete required to build the terrace.


Steps with a length of 50m long each


Ans: Given that a football ground comprises of \[15\] steps each of which is \[50\] m long and built of solid concrete. Each step has a rise of $\dfrac{1}{4}$m and a tread of $\dfrac{1}{2}$m. An easy illustration of the problem is depicted below.


Individual step with breadth and width


Here blue step is the lowermost step. Let it be known as step $1$. The volume of step $1$ is $\dfrac{1}{2}\times \dfrac{1}{4}\times 50\text{ }{{m}^{3}}$.

The red step is the second lowermost step. Let it be known as step $2$. The volume of step $2$ is $\dfrac{1}{2}\times \dfrac{1}{2}\times 50\text{ }{{m}^{3}}$.

The green step is the third lower step. Let it be known as step $3$. The volume of step $3$ is $\dfrac{1}{2}\times 1\times 50\text{ }{{m}^{3}}$.  


Steps with increasing height


We can see that the height is increasing with each increasing step by a factor of $\dfrac{1}{4}$, length and width being constant. Hence the volume of each step is increasing by $\dfrac{1}{2}\times \dfrac{1}{4}\times 50\text{ }{{m}^{3}}$.

Therefore, we can conclude that the volume of steps is in A.P. with first term and common difference both as $\dfrac{1}{2}\times \dfrac{1}{4}\times 50=\dfrac{25}{4}\text{ }{{m}^{3}}$.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, 

${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( \dfrac{25}{4} \right)+\left( \dfrac{25}{4} \right)\left( 15-1 \right) \right]$

$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( \dfrac{25}{4} \right)\left[ 16 \right]$ 

$\Rightarrow {{S}_{15}}=15\cdot 25\cdot 2$

$\therefore {{S}_{15}}=750$

Therefore, volume of concrete required to build the terrace is $750\text{ }{{m}^{3}}$.


Conclusion

In conclusion, Class 10 Maths Exercise 5.4 has provided a thorough understanding of finding the sum of the first n terms of an arithmetic progression. By learning to apply the sum formula, Sn=n/2(2a+(n−1)d), students have gained the ability to solve a variety of problems involving arithmetic progressions. Class 10 Exercise 5.4 has also demonstrated the practical applications of AP in solving real-life word problems. 


Class 10 Maths Chapter 5: Exercises Breakdown

Exercise

Number of Questions

Exercise 5.1

4 Questions & Solutions

Exercise 5.2

20 Questions & Solutions

Exercise 5.3

20 Questions & Solutions


CBSE Class 10 Maths Chapter 5 Other Study Materials


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

FAQs on NCERT Solutions for Class 10 Maths Chapter 5 - Arithmetic Progressions Exercise 5.4

1. What we are going to study in chapter 5?

NCERT class 10 maths arithmetic progression is part of unit 2 in algebra. In this chapter, students will also analyse the patterns that succeed in obtaining by adding a fixed number to the preceding terms. They'll even see about finding the nth and the sum of n consecutive terms. So this knowledge can be useful to solve some of the problems of everyday life. The chapter explores how to deal with the implementation of arithmetic progression with certain issues of daily life. There are also simple and easy problems, given at regular intervals, to ease the concepts.

2. Are the questions asked from Exercise 5.4 of Chapter 5 of Class 10 Maths in Board exams?

Exercise 5.4 of Chapter 5 of Class 10 Maths has been marked as an optional exercise and is not very important for the Class 10 board exams. But this does not mean that it is not important for your studies. This exercise is a summarization of the whole chapter, and if you solve these exercises, you will be able to solve all the questions that come from this chapter. If you want help solving this complex exercise, visit Vedantu’s official site (vedantu.com).

3. Which are the most important questions of Exercise 5.4 of Chapter 5 of Class 10 Maths?

All the questions that are mentioned in the NCERT textbook of Exercise 5.4 of Chapter 5 of Class 10 Maths are equally important for your studies. They help in strengthening your concepts. This exercise has several complicated questions which if you learn how to solve can help you in dealing with the topic in higher classes as well. For any additional help to be able to solve this exercise, you can get Vedantu’s NCERT Solutions for Exercise 5.4 of Chapter 5 of Class 10 Maths.

4. Does Vedantu provide solutions to all the questions of Exercise 5.4 of Chapter 5 of Class 10 Maths even though it is an optional exercise?

Exercise 5.4 of "Arithmetic Progressions" is a short exercise containing only 5 questions. You can find the solutions to all of these 5 questions on Vedantu's NCERT Solutions for Exercise 5.4 of Chapter 5 of Class 10 Maths. All the questions in this exercise are conceptual and thus may require more explanation.


Each question is explained in step-by-step detail by Vedantu's Mathematics experts.  This allows the students to strengthen the conceptual application of the topics taught in Chapter 5 "Arithmetic Progression."

5. What is Exercise 5.4 of Chapter 5 of Class 10 Maths based on?

Chapter 5 "Arithmetic Progression" Exercise 5.4 has been mentioned in the Class 10 NCERT textbook as an optional exercise. The exercise tests your understanding of the chapter on the topics of:

  1. Arithmetic Progressions

  2. nth Term of an AP

  3. The sum of the First n Terms of an AP

Although an optional exercise, practising the questions of Exercise 5.4 will strengthen your ability to apply the aforementioned concepts to any problem. This will provide excellent practice for CBSE Board Examinations.

6. How to download the NCERT Solutions for Exercise 5.4 of Chapter 5 of Class 10 Maths?

To download Vedantu's Exercise 5.4 Class 10 Maths NCERT Solutions, follow these fuss-free steps:

  1. Visit Vedantu’s website at Vedantu and choose Chapter 5.

  2. As you scroll down, you will find a blue button titled "Download PDF." Click on it.

  3. It will redirect you to a new page containing the link to begin your download promptly.

7. Can the sum formula be used to find the number of terms in an AP in Class 10 Maths Exercise 5.4?

Yes, if the sum​, the first term a and the common difference are known, you can rearrange the formula to solve for the number of terms n.

8. How can I solve word problems involving the sum of an AP?

According to Exercise 5.4 Class 10 Maths NCERT Solutions, translate the word problem into an algebraic expression by identifying the first term, common difference, and number of terms. Then, use the sum formula to find the required solution.

9. What are the key concepts to remember when solving problems in Exercise 5.4?

Key concepts include understanding the sum formula, correctly identifying the first term and common difference, and accurately substituting values into the formula.

10. What types of questions are included in Exercise 5.4?

Exercise 5.4 includes questions that require finding the nth term of an arithmetic progression, solving word problems using the nth term formula, and applying the concepts of AP in various scenarios.