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

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progression Exercise 5.4 are provided here to help the students solve the long answer questions provided in Exercise 5.4 of NCERT Class 10 Maths textbook. In order to solve long answer questions, students must get well-versed in these exercise-wise NCERT Class 10 Maths solutions.

Students often fail to solve long-answer mathematical problems. In order to clear the steps to solve long answer problems, students are advised to study this exercise solution thoroughly. By studying NCERT Solutions, they will be able to clear their doubts in solving long answer questions.

## Topics Covered in Exercise 5.4

1. Sum of First n Terms of an AP
2. Solved examples

• Help students determine the steps to solve long answer questions.
• Studying this NCERT Class 10 solution for Maths will help students score good marks in the examination.
• It helps students memorise different formulas and methodologies to solve arithmetic progression problems.

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

### Access Other Exercise Solutions of Class 10 Maths Chapter 5 – Arithmetic Progression

The other exercises of NCERT Class 10 Maths Chapter 5 can be accessed using the links below.

Exercise 5.1– 4 questions (1 MCQ and 3 descriptive type questions)
Exercise 5.2– 20 questions (1 fill in the blanks, 2 MCQs, 7 Short answer questions and 10 Long answer questions)
Exercise 5.3– 20 Questions (3 fill in the blanks, 4 daily life examples, and 13 descriptive-type questions)

### Access Answers to NCERT Class 10 Maths Chapter 5 – Arithmetic Progression Exercise 5.4

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term? [Hint: Find n for an < 0]

Solution:

Given, the AP series is 121, 117, 113, . . .,

Thus, the first term, a = 121

The common difference, d = 117-121= -4

By the nth term formula,

an = a+(n −1)d

Therefore,

an = 121+(n−1)(-4)

= 121-4n+4

=125-4n

To find the first negative term of the series, an < 0

Therefore,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Therefore, the first negative term of the series is the 32nd term.

2. The sum of the third and the seventh terms of an AP is 6, and their product is 8. Find the sum of the first sixteen terms of the AP.

Solution:

From the given statements, we can write

a3 + a7 = 6 …………………………….(i)

And

a3 ×a7 = 8 ……………………………..(ii)

By the nth term formula,

an = a+(n−1)d

Third term, a3 = a+(3 -1)d

a3 = a + 2d………………………………(iii)

And Seventh term, a7= a+(7-1)d

a7 = a + 6d ………………………………..(iv)

From equations (iii) and (iv), putting in equation(i), we get

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3–4d …………………………………(v)

Again, putting the eq.(iii) and (iv) in eq. (ii), we get

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

32 – 2d2 = 8

9 – 4d2 = 8

4d2 = 1

d = 1/2 or -1/2

Now, by putting both the values of d, we get

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2

We know that the sum of the nth term of AP is

Sn = n/2 [2a +(n – 1)d]

So, when a = 1 and d=1/2

Then, the sum of the first 16 terms is

S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2

Then, the sum of the first 16 terms is

S16 = 16/2 [2(5)+(16-1)(-1/2)] = 8(5/2)=20

3. A ladder has rungs 25 cm apart (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = -250/25 ]. Solution:

Given,

The distance between the rungs of the ladder is 25cm.

Distance between the top rung and bottom rung of the ladder is = = 5/2 ×100cm

= 250cm

Therefore, total number of rungs = 250/25 + 1 = 11

As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude that the rungs are decreasing in the order of AP.

And the length of the wood required for the rungs will be equal to the sum of the terms of the AP series formed.

So,

The first term, a = 45

The last term, l = 25

Number of terms, n = 11

Now, as we know, the sum of nth terms is equal to

Sn= n/2(a+ l)

Sn= 11/2(45+25) = 11/2(70) = 385 cm

Hence, the length of the wood required for the rungs is 385cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x, such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint :Sx – 1 = S49 – Sx ]

Solution:

Given,

Row houses are numbers from 1,2,3,4,5…….49.

Thus, we can see the houses numbered in a row are in the form of AP.

So,

The first term, a = 1

The common difference, d=1

Let us say the number of the houses can be represented as

Sum of nth term of AP = n/2[2a+(n-1)d]

Sum of number of houses beyond x house = Sx-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we can write

S49 – Sx = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the given condition, eq.(i) and eq(ii) are equal to each other.

Therefore,

x(x-1)/2 = 25(49) – x(x+1)/2

x = ±35

As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.

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 1 4 m and a tread of 1 2 m (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = ¼ ×1/2 ×50 m3.] Solution:

As we can see from the given figure, the first step is ½ m wide, 2nd step is 1m wide and 3rd step is 3/2m wide. Thus, we can understand that the width of the step by ½ m each time when the height is ¼ m. And also, given the length of the steps is 50m the time. So, the width of steps forms a series AP in such a way that

½ , 1, 3/2, 2, ……..

The volume of steps = Volume of Cuboid

Now,

The volume of concrete required to build the first step = ¼ ×1/2 ×50 = 25/4

The volume of concrete required to build the second step =¼ ×1×50 = 25/2

The volume of concrete required to build the second step = ¼ ×3/2 ×50 = 75/4

Now, we can see the volumes of concrete required to build the steps are in the AP series.

25/4 , 25/2 , 75/4 …..

Thus, applying the AP series concept,

The first term, a = 25/4

The common difference, d = 25/2 – 25/4 = 25/4

As we know, the sum of n terms is

Sn = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)

Upon solving, we get

Sn = 15/2 (100)

Sn=750

Hence, the total volume of concrete required to build the terrace is 750 m3.