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Last updated at May 29, 2018 by Teachoo

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Misc 13 If , (a+bx)/(aโbx) = (b+cx)/(bโcx) = (c+dx)/(aโdx) (x โ 0)then show that a, b, c and d are in G.P. Introduction Componendo dividendo If ๐ฅ/๐ฆ = ๐/๐ Applying componendo dividendo (๐ฅ + ๐ฆ)/(๐ฅ โ ๐ฆ) = (๐ + ๐)/(๐ โ ๐) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 โ 2) = (4 + 8)/(4 โ 8) 3/(โ1) = 12/(โ4) -3 = -3 Misc 13 If , (a+bx)/(aโbx) = (b+cx)/(bโcx) = (c+dx)/(aโdx) (x โ 0)then show that a, b, c and d are in G.P. We have (a+bx)/(aโbx) = (b+cx)/(bโcx) = (c+dx)/(c โ dx) & we want to show that a, b, c, d are in G.P. Taking (a+bx)/(aโbx) = (b+cx)/(bโcx) = (c+dx)/(c โ dx) Applying componendo dividendo (a + bx + a โ bx)/((a + bx) โ(aโbx)) = (b + cx + (b โ cx))/(b + cx โ(b โ cx)) = (c + dx + (c โ dx))/(c + dx โ (c โ dx)) (a + a + bx โ bx)/(๐๐ฅ+ bx โ a + a ) = (b + b + cx โ cx)/(cx + cx โ ๐ + ๐) = (c + dx + c โ dx)/(dx + dx โ ๐ + ๐) (2๐+0)/(2๐๐ฅ+0) = (2๐ + 0)/(2๐๐ฅ + 0) = (2๐+0)/(2๐๐ฅ+0) 2๐/2๐๐ฅ = 2๐/2๐๐ฅ = 2๐/2๐๐ฅ ๐/๐๐ฅ = ๐/๐๐ฅ = ๐/๐๐ฅ a/b " =" b/c = c/d b/a " =" c/b = d/c Thus, a, b, c & d are in GP because their common ratio is same

Miscellaneous

Misc 1

Misc 2

Misc 3 Important

Misc 4

Misc 5

Misc 6

Misc 7 Important

Misc 8

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13 You are here

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20

Misc 21 Not in Syllabus - CBSE Exams 2021

Misc 22

Misc 23 Not in Syllabus - CBSE Exams 2021

Misc 24 Not in Syllabus - CBSE Exams 2021

Misc 25 Important Not in Syllabus - CBSE Exams 2021

Misc 26

Misc 27

Misc 28 Important

Misc 29

Misc 30

Misc 31

Misc 32 Important

Chapter 9 Class 11 Sequences and Series

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.