These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.

Which term of the AP: 121, 117. 113, ….., is its first negative term?

Solution:

Given sequence is 121,117,113,…

It is an AP. with a = 121, d = 117 – 121 = – 4

Let nth term of this sequence be its first negative term.

So, a < 0

i.e., a + (n – 1 )d < 0

121 + (n – 1) – 4 < 0

121 – 4n + 4 < 0

125 < 4n ⇒ 4n > 125

⇒ n > \(\frac { 125 }{ 4 }\) = 31.25

Hence, 32^{th} term of the given sequence is its first negative term.

Question 2.

The sum of the third term and the seventh term of an AP is 6 and their product is 8. Fibd the sum of first sixteen terms of the AP?

Solution:

Third term of AP is a + (3 – 1 )d = a + 2d

Seventh term of AP is a + (7 – 1)d = a + 6d

We have,

Sum of third and seventh terms of AP

(a + 2d) + (a + 6 d) = 6

2a + 8d= 6

a + 4d= 3

⇒ a = 3 – 4d

Also given,

Product (a + 2d) (a + 6d)= 6

a² + 6ad + 2ad + 12d² = 8

a² + 8ad + 12d² = 8

Now put the value of a in equation, we get,

(3 – 4d)² + 8(3 – 4d) + 12d² = 8

⇒ a + 16d² – 24d + 24d – 32d² + 12d² = 8

a – 4d² = 8

⇒ 4d² = 1

⇒ d² = \(\frac { 1 }{ 4 }\)

So, d = ±\(\frac { 1 }{ 2 }\)

Taking d = ±\(\frac { 1 }{ 2 }\), Put the value of d in

Question 3.

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

Solution:

Length of the rungs of a ladder decreases uniformly from bottom to top.

Distance between bottom to top

= 2\(\frac { 1 }{ 2 }\)m

= \(\frac { 5 }{ 2 }\) x 100cm = 250cm

Hence, total no. of rungs = \(\frac { 250 }{ 25 }\) = 10

Series is 45,45 + d, 45 + 2d,… 45 + 9d

First term = a = 45, last term = 45 + 9d

According to question,

45 + 9d = 25

45 – 25 = 9d

20 = 9d

d = \(\frac { -20 }{ 9 }\)

Therefore, length of the wood required for rungs is,

Question 4.

The houses of a row are numbered consecutively from 1 to 49. Show that there is 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.

Solution:

The series of houses will be 1, 2, 3, 4, …… 49

S_{x-1} = S_{49} – S_{x}

Let x be the number

According to the given condition

Now, put the values of S_{x-1}, S_{49} and (S_{x-1} = S_{49} – S_{x})

Question 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 \(\frac { 1 }{ 2 } m\) and a tread of \(\frac { 1 }{ 2 } m\). Calculate the total volume of concrete required to build the terrace.

Solution:

Volume of concrete required to build the first step.

= \(\frac { 1 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³

Volume of concrete required to build the second step

= \(\frac { 2 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³

Similarly, third step = \(\frac { 3 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³

and for 15th = \(\frac { 15 }{ 4 }\) x \(\frac { 1 }{ 2 }\) x 50m³

So, total volume of concrete required to build 15 steps

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