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AP Board Class 10 Maths Chapter 6 Progressions Ex 6.5 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.5 Book Answers |
Andhra Pradesh Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Textbooks Solutions PDF
Andhra Pradesh State Board STD 10th Maths Chapter 6 Progressions Ex 6.5 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.5 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 10th Maths Chapter 6 Progressions Ex 6.5 Textbooks. These Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.Andhra Pradesh State Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Books Solutions
Board | AP Board |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 10th |
Subject | Maths |
Chapters | Maths Chapter 6 Progressions Ex 6.5 |
Provider | Hsslive |
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AP Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 10th Maths Chapter 6 Progressions Ex 6.5 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers
Question 1.
For each geometric progression find the common ratio ‘r’, and then find an.
i) 3, 32, 34, 38, …….
Answer:
Given G.P.: 3, 32, 34, 38, …….
ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = 𝑎2𝑎1 = −62 = -3
an = a . rn-1 = 2 × (-3)n-1
∴ r = -3; an = 2(-3)n-1
iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = 𝑎2𝑎1 = −3−1 = 3
an = a . rn-1 = (-1) × 3n-1
∴ r = 3; an = (-1) × 3n-1
iv) 5, 2, 45, 825, …….
Given G.P. = 5, 2, 45, 825, …….
a = 5, r = 𝑎2𝑎1 = 25
an = a . rn-1 = 5 × (25)𝑛−1
∴ r = 25; an = 5(25)𝑛−1
Question 2.
Find the 10th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = 𝑎2𝑎1 = 255 = 5
an = a . rn-1 = 5 × (5)n-1 = 51+n-1 = 5n
a10 = a . r9 = 5 × 59 = 510
∴ a10 = 510; an = 5n
Question 3.
Find the indicated term of each geometric progression.
i) a1 = 9; r = 13; find a7.
Answer:
an = a . rn-1
ii) a1 = -12; r = 13; find a6.
Answer:
an = a . rn-1
Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = 𝑎2𝑎1 = 82 = 4
Let the nth term of G.P. be 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 29
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = 102 = 5
∴ 512 is the 5th term of the given G.P.
ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = 𝑎2𝑎1 = 33√ = √3
now an = a . rn-1 = 729
⇒ (√3)(√3)n-1 = 729
⇒ (√3)n = 36 = (√3)12
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.
iii) 13, 19, 127, ……. is 12187?
Answer:
Given G.P.: 13, 19, 127, ……. is 12187
Let 12187 be the nth term of the G.P., then
an = a . rn-1 = 12187
[∵ bases are equal, exponents are also equal]
7th term of G.P is 12187.
Question 5.
Find the 12th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a8 = 192 and r = 2
an = a . rn-1
a8 = a . (2)8-1 = 192
= 3 × 210 = 3 × 1024 = 3072.
Question 6.
The 4th term of a geometric progression is 23 and the seventh term is 1681. Find the geometric series.
Answer:
Given: In a G.P.
Now substituting r = 23 in equation (1)
we get,
Question 7.
If the geometric progressions 162, 54, 18, ….. and 281, 227, 29,….. have their nth term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and 281, 227, 29,……
Given that nth terms are equal
an = a . rn-1
⇒ 3n-1+n-1 = 81 × 81
⇒ 32n-2 = 34 × 34
⇒ 32n-2 = 38 [∵ am . an = am+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = 102 = 5
The 5th terms of the two G.P.s are equal.
AP Board Textbook Solutions PDF for Class 10th Maths
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