AP Board Class 10 Maths Chapter 6 Progressions Ex 6.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.4 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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AP Board Class 10 Maths Chapter 6 Progressions Ex 6.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.4 Book Answers

AP Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.4 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.4 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 6 Progressions Ex 6.4 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 6 Progressions Ex 6.4
Provider	Hsslive

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AP Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Textbooks Solutions with Answer PDF Download

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10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved in the form a G.P.?
i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .
Answer:
Given: Sharmila’s yearly salary = Rs. 5,00,000.
Rate of annual increment = 10 %.

Here, a = a₁ = 5,00,000
a₂ = 5,00,000 × 1110 = 5,50,000
a₃ = 5,00,000 × 1110 × 1110 = 6,05,000
a₄ = 5,00,000 × 1110 × 1110 × 1110 = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number 1110.
∴ r = common ratio = 1110
Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Answer:
Given: Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks.
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are 100, 98, 96, 94, ….
Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:
Given: An equilateral triangle whose perimeter = 24 cm.
Side of the equilateral triangle = 243 = 8 cm.
[∵ All sides of equilateral are equal] ……. (1)
Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = 82 = 4 cm
[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]
Similarly, the side of third triangle = 42 = 2 cm
∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,
a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number 12.
∴ The situation forms a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.
i) a = 4 ; r = 3.
Answer:
The terms are a, ar, ar², ar³, ……..
∴ 4, 4 × 3, 4 × 3² , 4 × 3² , ……
⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = 15
Answer:
The terms are a, ar, ar², ar³, ……..

iii) a = 81 ; r = –13
Answer:
The terms of a G.P are:
a, ar, ar², ar³, ……..

⇒ 81, -27, 9,

iv) a = 164; r = 2.
Answer:
Given: a = 164; r = 2.

∴ The G.P is 164, 132, 116, …….

Question 3.
Which of the following are G.P. ? If they are G.P, write three more terms,
i) 4, 8, 16, ……
Answer:
Given: 4, 8, 16, ……
where, a₁ = 4; a₂ = 8; a₃ = 16, ……
𝑎2𝑎1=84=2
𝑎3𝑎2=168=2
∴ r = 𝑎2𝑎1=𝑎3𝑎2 = 2
Hence 4, 8, 16, … is a G.P.
where a = 4 and r = 2
a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32
a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64
a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) 13, –16, 112, …….
Answer:
Given: t₁ = 13, t₂ = –16, t₃ = 112, ….
13, –16, 112, …….

Hence the ratio is common between any two successive terms.
∴ 13, –16, 112, ……. is G.P.
where a = 13 and r = –12

iii) 5, 55, 555, ……..
Answer:
Given: t₁ = 5, t₂ = 55, t₃ = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……
Given: t₁ = -2, t₂ = -6, t₃ = -18

∴ -2, -6, -18, is a G.P.
where a = -2 and r = 3
a_n = a . r^n-1 =
a₄ = a . r³ = (-2) × 3³ = -2 × 27 = -54
a₅ = a . r⁴ = (-2) × 3⁴ = -2 × 81 = -162
a₆ = a . r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) 12, 14, 16, …….
Answer:

i.e., 12, 14, 16, ….. is not a G.P.

vi) 3, -3², 3³, ……
Given: t₁ = 3, t₂ = -3², t₃ = 3³, ……

i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, …… forms a G.P,
where a = 3; r = -3
a_n = a . r^n-1
a₄ = a . r³ = 3 × (-3)³ = 3 × (-27) = -81
a₅ = a . r⁴ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = a . r⁵ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, 1𝑥, …….
Answer:
Given: t₁ = x, t₂ = 1, t₃ = 1𝑥, ……

Hence x, 1, 1𝑥, …. forms a G.P.
where a = x; r = 1𝑥

viii) 12√, -2, 82√, …….
Answer:
Given: t₁ = 12√, t₂ = -2, t₃ = 82√, ……

ix) 0.4, 0.04, 0.004, ……..
Answer:
Given: t₁ = 0.4, t₂ = 0.04, t₃ = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.
where a = 0.4; r = 110

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Answer:
Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.
∴ r = t2t1 = t3t2
⇒ 𝑥+2𝑥 = 𝑥+6𝑥+2
⇒(x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = – 4 = -2x = -4
∴ x = 2

AP Board Textbook Solutions PDF for Class 10th Maths

AP Board Class 10 Maths Chapter 1 Real Numbers Textbook Solutions PDF

AP Board Class 10 Maths Chapter 2 Sets Textbook Solutions PDF

AP Board Class 10 Maths Chapter 3 Polynomials Textbook Solutions PDF

AP Board Class 10 Maths Chapter 4 Pair of Linear Equations in Two Variables Textbook Solutions PDF

AP Board Class 10 Maths Chapter 5 Quadratic Equations Textbook Solutions PDF

AP Board Class 10 Maths Chapter 6 Progressions Textbook Solutions PDF

AP Board Class 10 Maths Chapter 7 Coordinate Geometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 8 Similar Triangles Textbook Solutions PDF

AP Board Class 10 Maths Chapter 9 Tangents and Secants to a Circle Textbook Solutions PDF

AP Board Class 10 Maths Chapter 10 Mensuration Textbook Solutions PDF

AP Board Class 10 Maths Chapter 11 Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 12 Applications of Trigonometry Textbook Solutions PDF

AP Board Class 10 Maths Chapter 13 Probability Textbook Solutions PDF

AP Board Class 10 Maths Chapter 14 Statistics Textbook Solutions PDF

Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.4 Textbooks for Exam Preparations

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Thursday, June 9, 2022