KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 1.
Which term of the A.P: 121, 117, 113,……, is its first negative term?
[Hint: Find n for an < 0]
Answer:
The given AP is 121, 117, 113,…….
a = 121, d = 117 – 121 = – 4
Let the nth term of the AP be the first negative term.
∴ an < 0
a + ( n – 1 )d < 0
121 + (n – 1) (- 4) < 0
121 – 4n + 4 < 0
125 < 4n
4n > 125
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 1
∴ n = 32.
Hence, 32nd term of the given AP is the first negative term.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

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:
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q2
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 3
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 4

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 3.
A ladder has rungs 25 cm apart. (see Fig.1.7 ). The rungs decrease uniformly ¡n length form 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].

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 5
Answer:
Distance between top and bottom rung
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 6
Distance between every two rungs
= 25 cm [given]
Number of rungs (n) = + 1 for top most rung]
= 10 + 1 = 11
∴ n = 11
Because the rungs decrease uniformly in length from 45 cm at the bottom to 25cm at the top
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 7

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 number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: Sx-1 = S49 – Sx]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q4
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 10
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q4 1

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

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 (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 = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{m}^{3}\)
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 20
Answer:
Volume of concrete required to build the first step length × breadth × Height.
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 21
∴ Total volume of concrete required to build the terrace of 15 steps
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 22
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

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