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# NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progression - Exercise 5.4

Last updated date: 19th Jun 2024
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## NCERT Solutions Class 10 Maths Arithmetic Progression Chapter 5

 Class: NCERT Solutions for Class 10 Subject: Class 10 Maths Chapter Name: Chapter 5 - Arithmetic Progressions Exercise: Exercise - 5.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progression - Exercise 5.4
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## Access NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progression

### 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 $=\dfrac{250}{25}$)

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., ${{S}_{x-1}}={{S}_{49}}-{{S}_{x}}$

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.

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.

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 $\dfrac{1}{2}\times 1\times 50\text{ }{{m}^{3}}$.

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 $750\text{ }{{m}^{3}}$.

## Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.4 - Free PDF Download

NCERT Solution Class 10 Maths includes Mathematics Book textbook solutions. NCERT Solutions has a total of 15 chapters for CBSE Class 10 Maths. On our website, you can download Class 10 Maths NCERT Solutions in PDF for free. The NCERT Maths class 10 PDF with the latest changes are available on Vedantu's app and website as per the latest CBSE syllabus.

To gain a deeper understanding of exercise questions, download NCERT Solutions for Class 10 Mathematics Chapter 5. You'll be able to take part in free conceptual videos and live masterclasses with the Vedantu learning app. You will also have access to all free PDFs for solutions and study materials.

Download the Class 10 Maths Free NCERT Solutions Chapter 5 Exercise 5.4 PDF. NCERT Solutions are highly helpful while doing your homework. The experienced Teachers prepare all the answers for Class 10 Maths. Detailed answers to all the questions in the NCERT TextBook Chapter 5, Maths Class 10 Arithmetic Progressions Exercise 5.4.

Practicing various types of questions will ensure that you understand the topic after studying Chapter 5. Free PDFs of NCERT Solutions are also available here and can be downloaded free of charge for online and offline use. Download solutions for NCERT now. These NCERT Solutions for Class 10 Mathematics Chapter 3 will help you plan effectively for your board’s study. Expert teachers design the NCERT solutions for Maths Class 10 Chapter 5 exercise 5.4. The concepts are explained in depth in NCERT Solutions for Class 10 Maths Chapter 5 and all doubts are immediately resolved by our subject matter experts during live doubt-solving sessions. You can also understand all the challenging concepts by referring to NCERT's class 10 maths chapter 5 solutions and score good marks in your exams.

### Benefits of NCERT Solutions for Class 10 Maths

In order to make your learning process easier, our professional and trained teachers have formulated all the solutions in a well-structured format. The following are the benefits that you would have if you use our free Class 10 Chapter 5 Maths Solutions:

• All solutions available are reliable and of high quality.

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## FAQs on NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progression - 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 analyze 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. How to reduce the fear of Maths?

Some of the students face challenges while they are learning Maths. Thus, with the aid of its expert teachers, Vedantu has made the subject easier to understand. The best way to learn is to take a smart move forward and download the NCERT Solutions PDF now. Vedantu's NCERT Solutions is one of the most significant sections of Class 10 Maths study materials.

These solutions have been crafted with the utmost care by qualified and professional teachers to make your preparation for the exams easier.

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

The 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).

4. 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

5. 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."

6. 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. Sum of 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.

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