Well-designed AP 10th Class Maths Textbook Solutions Chapter 5 Arithmetic Progressions Exercise 5.4 offers step-by-step explanations to help students understand problem-solving strategies.

## Arithmetic Progressions Class 10 Exercise 5.4 Solutions – 10th Class Maths 5.4 Exercise Solutions

Question 1.

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

[Hint : Find n for a_{n} < 0]

Solution:

Given A.P is 121, 117, 113,………

a = 121, d = 117 – 121 = – 4

We know that,

a_{n} = a + (n – 1) d

= 121 + (n – 1) (- 4)

a_{n} = 121 – 4n + 4

= 125 – 4n = 0

If a_{n} = 0, then 125 – 4n = 0

4n = 125

n = \(\frac{125}{4}\) = 31 terms

If a_{n} < 0 means n > 31 that is n = 32, 33,……

If n = 32, then a_{n} = 125 – 4n

a_{32} = 125 – 4(32) = 125 – 128 = -3

At n = 32, a_{n} will become the first negative number.

Question 2.

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

Solution:

Let the terms in A.P are a, a + d, a + 2d,…

We know that, a_{n} = a + (n – 1) d

a_{3} = a + 2d, a_{7} = a + 6d

a_{3} + a_{7} = a + 2d + a + 6d = 6

⇒ 2a + 8d = 6

⇒ 2 (a + 4d) = 6

⇒ a + 4d = 3 → (1)

a_{3} × a_{7} = 8

⇒ (a + 2d) (a + 6d) = 8

⇒ (a + 4d – 2d) (a + 4d + 2d) = 8 → (2)

Put (1) in (2) ⇒ (3 – 2d) (3 + 2d) = 8

⇒ 3^{2} – (2d)^{2} = 8

⇒ 9 – 4d^{2} = 8

⇒ -4d^{2} = 8 – 9

⇒

If d = \(\frac{1}{2}\), then a + 4d = 3

⇒

⇒ a + 2 = 3

∴ a = 3 – 2 = 1

If d = –\(\frac{1}{2}\), then

⇒ a – 2 = 3

∴ a = 3 + 2 = 5

If a = 1, d = +\(\frac{1}{2}\) and n = 16

∴ 16^{th} terms are 20 and 76.

Question 3.

A ladder has rungs 25 cm apart. 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 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]

Solution:

Given, the distance between the two consecutive rungs = 25 cm

Length of the bottom rung = 45 cm

Length of the last rung = 25 cm

Lengths of rungs are in A.P : 45,… 25 cm (11 rungs)

Number of rungs = \(\frac{2.5 \mathrm{~m}}{25 \mathrm{~cm}}\) + 1 = \(\frac{250}{25}\) + 1

= 10 + 1 = 11

a = 45, n = 11, l = 25 cm

We know that S_{n} = \(\frac{\mathrm{n}}{2}\) (a + l)

= \(\frac{11}{2}\) (45 + 25)

Total length of the required wood = 385 cm (or) 3.85 m.

Question 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: S_{x – 1} = S_{49} – S_{x}]

Solution:

Let there be 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 as:

House : H_{1}H_{2}H_{3},…H_{x-1}H_{x}…H_{x+1}..H_{49}

Number : 1, 2, 3,… x – 1, x, x + 1,… 49

Sum = 1 + 2 + 3 + … + (x – 1) + x + (x + 1) + …….. + 49

Sum of 1 to (x – 1) numbers = Sum of (x + 1) to 49 numbers.

⇒ 1 + 2 + 3 + …… + (x – 1) = (x + 1) + (x + 2) + (x + 3) + ……….. + 49

⇒ [(1 + 2 + 3 + …….. + x) + (x + 1) + (x + 2) +……. + 49] – [1 + 2 + 3 + ……. + x]

We know that,

So, 35^{th} house is the value of x satisfying the given condition exists.

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 \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m. Calculate the total volume of concrete required to build the terrace.

[Hint: Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m^{3}]

Solution:

Given number of steps = 15

Length of each step = 50 m

Breadth of step = \(\frac{1}{2}\) m

Height of first step = \(\frac{1}{2}\) m

Height of the each step is increased by \(\frac{1}{2}\) m.

Let V_{1}, V_{2}, V_{3},. . ….. V_{15} are the volumes of each step.

Total volume S_{15} = V_{1} + V_{2} + V_{3} +……. + V_{15}

V = l ∙ b ∙ h

Total volume of concrete required to build a terrace is 750 m^{3}.