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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 8 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.