Misc 13 - If (a + bx)/(a - bx) = (b + cx)/(b-cx) = (c + dx) - Miscellaneous

Misc 13 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 13 - Chapter 9 Class 11 Sequences and Series - Part 3

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Misc 13 If , (a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(a−dx) (x ≠ 0)then show that a, b, c and d are in G.P. Introduction Componendo dividendo If 𝑥/𝑦 = 𝑎/𝑏 Applying componendo dividendo (𝑥 + 𝑦)/(𝑥 − 𝑦) = (𝑎 + 𝑏)/(𝑎 − 𝑏) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 − 2) = (4 + 8)/(4 − 8) 3/(−1) = 12/(−4) -3 = -3 Misc 13 If , (a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(a−dx) (x ≠ 0)then show that a, b, c and d are in G.P. We have (a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(c − dx) & we want to show that a, b, c, d are in G.P. Taking (a+bx)/(a−bx) = (b+cx)/(b−cx) = (c+dx)/(c − dx) Applying componendo dividendo (a + bx + a − bx)/((a + bx) −(a−bx)) = (b + cx + (b − cx))/(b + cx −(b − cx)) = (c + dx + (c − dx))/(c + dx − (c − dx)) (a + a + bx − bx)/(𝑏𝑥+ bx − a + a ) = (b + b + cx − cx)/(cx + cx − 𝑏 + 𝑏) = (c + dx + c − dx)/(dx + dx − 𝑐 + 𝑐) (2𝑎+0)/(2𝑏𝑥+0) = (2𝑏 + 0)/(2𝑐𝑥 + 0) = (2𝑐+0)/(2𝑑𝑥+0) 2𝑎/2𝑏𝑥 = 2𝑏/2𝑐𝑥 = 2𝑐/2𝑑𝑥 𝑎/𝑏𝑥 = 𝑏/𝑐𝑥 = 𝑐/𝑑𝑥 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.