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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 Safeer PP | Updated on Jul 06, 2022 03:50 PM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Maths exercise 5.4 – 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 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.

This Story also Contains
  1. Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.4
  2. More About NCERT Solutions for Class 10 Maths Exercise 5.4
  3. Benefits of NCERT Solutions for Class 10 Maths Exercise 5.4
  4. NCERT Solutions Subject Wise
  5. Subject wise NCERT Exemplar Solutions

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.

The general form of an AP is 1639373147483 , 1639373146589 , 1639373146890 , …

The other formulas to Arithmetic progression are:

For finding nth term: 1639373148549 1639373147189

{where is the 1639373147950 common difference and 1639373148850 is the first term}

Sum of first n terms: 1639373149228

{where is the 1639373148107 common difference and 1639373149054 is the first term}

NCERT book Exercise 5.4 Class 10 Maths an optional exercise with a total of 5 questions that may appear difficult at first and will necessitate some brainstorming.

Along with NCERT syllabus Class 10 Maths chapter 5 exercise 5.4 the following exercises are also present.

Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.4

Q1 Which term of the AP: is its first negative term? [ Hint : Find n for a_n<0 ]

Answer:

Given AP is
\small 121,117,113,...,
Here a = 121 \ and \ d = -4
Let suppose nth term of the AP is first negative term
Then,
a_n = a+ (n-1)d
If 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, first negative term must be 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 sum of third and seventh terms of an AP are 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)
put value from equation (i) in (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 bottom-most rungs is of 45 cm length and topmost is of 25 cm length.As it is given that the length of rungs decrease 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, 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 \small 1 to \small 49 . 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 this value of \small x . [ Hint : \small S_{x-1}=S_4_9-S_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 their exist an n term such that ( n < 49)
Now, according to given conditions
Sum of first n - 1 terms of AP = Sum of terms following the nth term
Sum of first n - 1 term of AP = Sum of whole AP - Sum of 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 number are not negative so we reject n = -35

Therefore, the sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 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 clearly 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 step 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

More About NCERT Solutions for Class 10 Maths Exercise 5.4

NCERT solutions Class 10 Maths chapter 5 exercise 5.4 – This exercise features questions based on realistic real-life scenarios. One question, for example, is based on the ladder notion, and students must compute the length of the wood. Solving this optional exercise 5.4 Class 10 Maths will help us grasp the concept of arithmetic progression deeply. It will also be helpful for us to use the formula to get the sum of the first ‘n’ number of terms or to get a particular term of an arithmetic series. Students should practise many variants of this question because problems like this are frequently asked in board exams.

Also Read| Arithmetic Progressions Class 10 Notes

Benefits of NCERT Solutions for Class 10 Maths Exercise 5.4

  • If you go through Class 10 Maths chapter 5 exercise 5.4, you will be able to get more marks if you practise it thoroughly.

  • Class 10 Maths chapter 5 exercise 5.4 is based on practical use of Arithmetic Progression and its Use.

  • NCERT solutions for Class 10 Maths chapter 5 exercise 5.4 helps in solving and revising all questions which are related to the previous exercises.

  • If you go through Class 10 Maths chapter 5 exercise 5.4, we can see that the questions are a bit tough, this will help in solving questions that are of higher standards and are difficult.

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Frequently Asked Question (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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