RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.4

Question 1.
Which term of the AP : 121, 117, 113, ……. is its first negative term?
[Hint: Find n for a_n < 0]
Solution:
According to the question AP : 121, 117, 113, …….
Here, a = a₁ = 121, a₂ = 117, a₃ = 113,
d = a₂ – a₁ = 117 – 121 = -4
∵ a_n = a + (n – 1) d
a_n = 121 + (n – 1) (- 4)
= 121 – 4n + 4= 125 – 4n
a_n < 0
or 125 – 4n < 0 or 4n > 125
or 125 > 4n
or n > \(\frac{125}{4}\) or n > 31\(\frac{1}{4}\)
or n > 31.25
or n < 32
Since n is an integer.
But for the first negative term n will be an integer.
∴ n = 32
Hence, 32nd term of the given AP will be the first negative term.

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 ‘a’ and ‘d’ be respectively the first term and the common difference of the given AP.
According to the first condition,
a₃ + a₇ = 6
⇒ [a + (3 – 1) d] + [a + (7 – 1) d] = 6
[∵ a_n = a + (n – 1) d]
or a + 2d + a + 6d = 6
or 2a + 8d = 6
or a + 4d = 3 ….(1)
According to the second condition,
a₃ (a₇) = 8
⇒ [a + (3 – 1) d] [a + (7 – 1) d] = 8
or (a + 2d) (a + 6d) = 8
or [3 – 4d + 2d] [3 – 4d + 6d] = 8
[From (1), a = 3 – 4d]
or (3 – 2d) (3 + 2d) = 8
or 9 – 4d² = 8
or 4d² = 9 – 8
or d² = \(\frac{1}{4}\)
or d = ±\(\frac{1}{2}\)
Case I: When d = \(\frac{1}{2}\)
Substituting the value d = \(\frac{1}{2}\) in (1)
a + 4(\(\frac{1}{2}\)) = 3
or a + 2 = 3 or a = 3 – 2 = 1
Using the formula,
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
∴ S₁₆ = \(\frac{16}{2}\)[2(1) + (16 – 1)\(\frac{1}{2}\)]
= 8[2 + \(\frac{15}{2}\)]
= 8 [\(\frac{4+15}{2}\) = \(\frac{19}{2}\)]
S₁₆ = 76

Case II: When d = \(-\frac{1}{2}\)
Substituting d = \(-\frac{1}{2}\) in (1),
a + 4(\(-\frac{1}{2}\)) = 3
or a – 2 = 3
or a = 3 + 2 = 5
Using the formula
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
S₁₆ = \(\frac{16}{2}\)[2(5) + (16 – 1)\(-\frac{1}{2}\)]
= 8[10 – \(\frac{15}{2}\)]
= 8 [\(\frac{20-15}{2}\) = \(\frac{5}{2}\)]
S₁₆ = 20
Hence sum of 16 terms = 20

Question 3.
A ladder has rungs 25 cm apart (see Fig.). 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:
Distance between the first and the last rung
= 2\(\frac{1}{2}\)m = \(\frac{5}{2}\) m
= (\(\frac{5}{2}\) × cm = 250 cm
and the distance between two consecutive rungs = 25 cm
∴Number of rungs in the ladder
Distance between the first
= \(\frac{and the last rung}{Distance between two}\) + 1
consecutive rungs
= \(\frac{250}{25 }\) + 1 = 10 + 1 = 11
∵ Length of the first rung (a) = 25 cm
and length of the last rung (l) = 45 cm
∴ The total length of the wood used in 11 rungs
= \(\frac{n}{2}\) (a + l) = \(\frac{11}{2}\) (25 + 45)
= \(\frac{11}{2}\) × 70= 11 × 35
= 385 cm = 3.85 m
Hence, the length of the wood used in the rungs of the ladder = 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 = S₄₉ – S_x]
Solution:Let ‘x’ represent the number of any house.
Here a = a₁ = 1; d = 1
According to the question

Hence the value ofx is 35.

Question 5.
A small terrace at a football ground comprises of IS 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. (see Fig.). 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 m³]

Solution:
Volume of the concrete used in the first step
= (\(\frac{1}{4}\) × \(\frac{1}{2}\) × 50)m³
= (\(\frac{25}{4}\))m³
Volume of the concrete used in the second step
= (\(\frac{2}{4}\) × \(\frac{1}{2}\) × 50)m³
= (\(\frac{25}{2}\)) m³
Volume of the concrete used in the third step
= (\(\frac{3}{4}\) × \(\frac{1}{2}\) × 50)m³
= (\(\frac{75}{4}\))m³
and so on so forth upto 15 steps
Here, a = a₁ = \(\frac{25}{4}\); a₂ = \(\frac{25}{2}\); a₃ = \(\frac{75}{4}\)
and n = 15

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