# Misc 20

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 20 If a, b, c are in A.P, ; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P. It is given that a, b, c are in AP So, their common difference is same b – a = c – b b + b = c + a 2b = c + a b = (𝑐 + 𝑎)/2 Also given that b, c, d are in GP So, their common ratio is same 𝑐/𝑏 = 𝑑/𝑐 c2 = bd Also 1/c, 1/d, 1/e are in A.P. So, their common difference is same 1/d − 1/c= 1/e− 1/d 1/d + 1/d = 1/e + 1/c 2(1/d) = (c + e)/ec 2/d = (c + e)/ec d/2 = ec/(c + e) d = 2(ec/(c+e)) We need to show that a, c, e are in GP i.e. we need to show their common ratio is same c/a = e/c c2 = ae So, we need to show c2 = ae From (2), we have c2 = bd Putting value of b = (𝑎 + 𝑐)/2 & d = 2𝑐𝑒/(𝑐 + 𝑒) c2 = ((a + c)/2)(2ce/(c + e)) c2 = ((a + c)(2ce))/(2(c + e)) c2 = ((a + c)(ce))/((c + e)) 𝑐2/𝑐 = (𝑒(𝑎 + 𝑐))/(𝑐 + 𝑒) c = (𝑒(𝑎 + 𝑐))/(𝑐 + 𝑒) c(c + e) = e(a + c) c2 + ec = ea + ec c2 = ea + ec – ec c2 = ea + 0 c2 = ea Which is what we need to prove Hence proved Thus, a, c & e are in GP

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

Misc 14

Misc 15

Misc 16 Important

Misc 17

Misc 18

Misc 19 Important

Misc 20 You are here

Misc 21

Misc 22

Misc 23

Misc 24

Misc 25 Important

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Misc 28 Important

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Misc 32 Important

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.