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

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

Edited By Komal Miglani | Updated on Apr 29, 2025 05:36 PM IST | #CBSE Class 10th

A progression is a sequence (or a set) of numbers that follow a specific pattern of repetition. We can define Arithmetic Progression as a set of numbers where the difference between any two consecutive terms remains the same throughout the series. This difference between the two terms is called the common difference of the AP. The general form of an AP is a, a + d, a + 2d, .... , where d is the common difference, and a is the first term.

This Story also Contains
  1. Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.4
  2. Topics Covered in Chapter 5, Arithmetic Progression: Exercise 5.4
  3. NCERT Solutions of Class 10 Subject Wise
  4. NCERT Exemplar Solutions of Class 10 Subject Wise

A brief summary is also provided at the end of this NCERT Solutions Class 10 Mathematics chapter 5 exercise 5.4, which will aid students in quickly memorising the full chapter as well as the formulas. NCERT Book Exercise 5.4 Class 10 Maths is an optional exercise with a total of 5 questions that may appear difficult at first and will necessitate some brainstorming.

Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.4

Q1. Which term of the AP: is its first negative term $\small 121,117,113,...,$? [Hint : Find $n$ for $a_n<0$]

Answer:

Given AP is

$\small 121,117,113,...,$

Here $a = 121 \ and \ d = -4$

Let's suppose the nth term of the AP is the first negative term

Then,

$a_n = a+ (n-1)d$

If the nth term is negative, then $a_n < 0$

$\Rightarrow 121+(n-1)(-4) < 0$

$\Rightarrow 125<4n$

$\Rightarrow n > \frac{125}{4}=31.25$

Therefore, the first negative term must be the 32nd term

Q2. The sum of the third and the seventh terms of an AP is $\small 6$ and their product is $\small 8$ . Find the sum of first sixteen terms of the AP.

Answer:

It is given that the sum of the third and seventh terms of an AP is and their product is $\small 8$

$a_3= a+ 2d$

$a_7= a+ 6d$

Now,

$a_3+a_7= a+ 2d+a+6d= 6$

$\Rightarrow 2a+8d = 6$

$\Rightarrow a+4d = 3 \Rightarrow a = 3-4d \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

And

$a_3.a_7 = (a+2d).(a+6d)=a^2+8ad +12d^2 = 8 \ \ \ \ \ \ \ -(ii)$

Putting the value from equation (i) in equation (ii), we will get

$\Rightarrow (3-4d)^2+8(3-4d)d+12d^2= 8$

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

$\Rightarrow 4d^2 = 1$

$\Rightarrow d = \pm \frac{1}{2}$

Now,

Case (i) $d = \frac{1}{2}$

$a= 3 - 4 \times \frac{1}{2} = 1$

Then,

$S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\frac{1}{2} \right \}$

$S_{16}=76$

Case (ii) $d = -\frac{1}{2}$

$a= 3 - 4 \times \left ( -\frac{1}{2} \right ) = 5$

Then,

$S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\left ( -\frac{1}{2} \right ) \right \}$

$S_{16}=20$

Q3. A ladder has rungs $\small 25$ cm apart. (see Fig. $\small 5.7$ ). The rungs decrease uniformly in length from $\small 45$ cm at the bottom to $\small 25$ cm at the top. If the top and the bottom rungs are $\small 2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint:Number of rungs = $\frac{250}{25}+1$]

1635921645173

Answer:

It is given that

The total distance between the top and bottom rung $= 2\frac{1}{2}\ m = 250cm$

Distance between any two rungs = 25 cm

Total number of rungs = $\frac{250}{25}+1= 11$

And it is also given that the bottom-most rung is of 45 cm length and the top-most is of 25 cm length. As it is given that the length of rungs

decreases uniformly, it will form an AP with $a = 25, a_{11} = 45 \ and \ n = 11$

Now, we know that

$a_{11}= a+ 10d$

$\Rightarrow 45=25+10d$

$\Rightarrow d = 2$

Now, the total length of the wood required for the rungs is equal to

$S_{11} = \frac{11}{2}\left \{ 2\times 25+(11-1)2 \right \}$

$S_{11} = \frac{11}{2}\left \{ 50+20 \right \}$

$S_{11} = \frac{11}{2}\times 70$

$S_{11} =385 \ cm$

Therefore, the total length of the wood required for the rungs is equal to 385 cm

Q4 The houses of a row are numbered consecutively from $\ small1$ to $\ small49$. Show that there is a value of $\small x$ such that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the sum of the numbers of the houses following it. Find the value of $\ small x$.

Answer:

It is given that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the sum of the numbers of the

houses following it

And 1,2,3,.....,49 form an AP with a = 1 and d = 1

Now, we know that

$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$

Suppose there exists an n term such that ( n < 49)

Now, according to the given conditions

The sum of the first n - 1 terms of AP = The sum of the terms following the nth term

Sum of the first n - 1 terms of AP = Sum of the whole AP - Sum of the first m terms of AP

i.e.

$S_{n-1}=S_{49}-S_n$

$\frac{n-1}{2}\left \{ 2a+((n-1)-1)d \right \}=\frac{49}{2}\left \{ 2a+(49-1)d \right \}-\frac{n}{2}\left \{ 2a+(n-1)d \right \}$

$\frac{n-1}{2}\left \{ 2+(n-2) \right \}=\frac{49}{2}\left \{ 2+48 \right \}-\frac{n}{2}\left \{ 2+(n-1) \right \}$

$\frac{n-1}{2}\left \{ n\right \}=\frac{49}{2}\left \{ 50 \right \}-\frac{n}{2}\left \{ n+1 \right \}$

$\frac{n^2}{2}-\frac{n}{2}=1225-\frac{n^2}{2}-\frac{n}{2}$

$n^2 = 1225$

$n = \pm 35$

Given House numbers are not negative,e so we reject n = -35

Therefore, the sum of no of houses preceding house no 35 is equal to the sum of no of houses following house number 35.

Q5. A small terrace at a football ground comprises of $\small 15$ steps each of which is $\small 50$ m long and built of solid concrete. Each step has a rise of $\small \frac{1}{4}\: m$ and a tread of $\small \frac{1}{2}\: m$ . (see Fig. $\small 5.8$ ). Calculate the total volume of concrete required to build the terrace.
[ Hint: Volume of concrete required to build the first step $\small =\frac{1}{4}\times \frac{1}{2}\times 50\: m^3$ ]

1635921665136

Answer:

It is given that

football ground comprises of $\small 15$ steps each of which is $\small 50$ m long and Each step has a rise of $\small \frac{1}{4}\: m$

and a tread of $\small \frac{1}{2}\: m$

Now,

The volume required to make the first step = $\frac{1}{4}\times \frac{1}{2}\times 50 = 6.25 \ m^3$

Similarly,

The volume required to make 2nd step = $\left ( \frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{1}{2}\times \frac{1}{2}\times 50 = 12.5 \ m^3$

And

The volume required to make 3rd step = $\left ( \frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{3}{4}\times \frac{1}{2}\times 50 = 18.75 \ m^3$

And so on

We can see that this is an AP with $a= 6.25 \ and \ d = 6.25$

Now, the total volume of concrete required to build the terrace of 15 such steps is

$S_{15} =\frac{15}{2}\left \{ 2 \times 6.25 +(15-1)6.25 \right \}$

$S_{15} =\frac{15}{2}\left \{ 12.5 +87.5\right \}$

$S_{15} =\frac{15}{2}\times 100$

$S_{15} =15\times 50 = 750$

Therefore, the total volume of concrete required to build the terrace of 15 such steps is $750 \ m^3$

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Topics Covered in Chapter 5, Arithmetic Progression: Exercise 5.4

  1. Finding the nth Term of an AP: The nth term of an AP can be determined using the formula (an) = a + (n - 1)d, where 'a' is the first term, 'n' is the term number, and 'd' is the common difference.
  2. Solving Word Problems Using AP: These word problems are modelled using AP, by extracting the word problems and applying the appropriate formula to find the terms, the sum of related values.
  3. Finding the First Negative Term in an AP: In an AP, there is a common difference between given terms, and if there is a negative difference between given terms, then the smallest value for the 'n' term is negative.
  4. Using AP in Practical Calculations: AP is also used in various scenarios such as finding the distance, model pattern of growth, seating arrangement, business calculations, etc.
  5. Advanced-Level Questions: In this exercise, 5.4, many advanced-level questions are included that can not be solved directly by applying the AP formulas. To solve these problems, students need to understand the underlying concept and apply it accordingly.
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NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.


Frequently Asked Questions (FAQs)

1. What is an Arithmetic progression?

We can define Arithmetic Progression as a set of collective numbers where the difference between any two consecutive terms remains the same throughout the series. This difference between the two terms is called the common difference of the AP.

2. What is the common difference of an Arithmetic progression?

The differences between any two consecutive terms are the same which is known as the common difference of the arithmetic progression.

3. Can common difference of an Arithmetic progression be negative?

Yes, it the Arithmetic progression is a decreasing one. For example:

15, 12, 9, 6, 3

4. Find the common difference of the arithmetic progression? 1, 4, 7, 10

Common difference =7-4=3

5. What is the formula for nth term of an AP

nth term=a+(n-1)d

Where 'a' is the first term and 'd' is the common difference.

6. What is the sum of first 10 natural numbers?

First ten natural numbers are 1, 2, 3, .........,10

This is ap with a common difference of 1. 

Sum =0.5n(first term+ last term)

=0.5 x 10(1+10)=5 x11=55

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Questions related to CBSE Class 10th

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Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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