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Last updated at May 29, 2018 by Teachoo
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Misc 17 If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P. We know that a, ar , ar2 , ar3, …. are in G.P. with first term a & common ratio r Given a, b, c, d are in G.P. So, a = a b = ar c = ar2 d = ar3 We want to show that (an + bn), (bn + cn), (cn + dn) are in GP i.e. to show common ratio are same (𝑏^𝑛+𝑐^𝑛)/(𝑎^𝑛+𝑏^𝑛 ) = (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Need to show (𝑏^𝑛+𝑐^𝑛)/(𝑎^𝑛+𝑏^𝑛 ) = (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Taking L.H.S (𝑏^𝑛+𝑐^𝑛)/(𝑎^𝑛+𝑏^𝑛 ) Putting b = ar , c = ar2 = (〖(𝑎𝑟)〗^𝑛+〖(〖𝑎𝑟〗^2)〗^𝑛)/(𝑎^𝑛+〖(𝑎𝑟)〗^𝑛 ) = (𝑎^𝑛 𝑟^𝑛 + 𝑎^𝑛.𝑟^2𝑛)/(𝑎^𝑛 + 𝑎^𝑛 𝑟^𝑛 ) = (𝑎^𝑛 𝑟^𝑛 (1 +𝑟^𝑛))/(𝑎^𝑛 (1 +𝑟^𝑛)) = rn Taking R.H.S (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Putting c = ar2, d = ar3, b = ar = (〖(〖𝑎𝑟〗^2)〗^𝑛 + 〖(〖𝑎𝑟〗^3)〗^𝑛)/(〖(𝑎𝑟)〗^𝑛 +〖 (〖𝑎𝑟〗^2)〗^𝑛 ) = (𝑎^𝑛 𝑟^2𝑛+ 𝑎^𝑛 .𝑟^2𝑛)/(𝑎^𝑛 𝑟^𝑛+〖 𝑎〗^𝑛 𝑟^2𝑛 ) = (𝑎^𝑛 (𝑟^2𝑛 + 𝑟^3𝑛))/(𝑎^𝑛 (𝑟^𝑛 + 𝑟^2𝑛)) = ( (𝑟^2𝑛 + 𝑟^3𝑛))/((𝑟^𝑛 + 𝑟^2𝑛)) = (𝑟^2𝑛 (1 + 𝑟))/(𝑟^𝑛 (1 + 𝑟)) = 𝑟^2𝑛/𝑟^𝑛 = (𝑟^𝑛.𝑟^𝑛)/𝑟^𝑛 = rn = rn = L.H.S Thus L.H.S = R.H.S Hence (an + bn), (bn + cn) & (cn + bn) are in GP Hence proved
Miscellaneous
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 You are here
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
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