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Misc 17 - If a, b, c, d are in GP, prove (an + bn), (bn + cn) - Geometric Progression(GP): Calculation based/Proofs

Misc 17 - Chapter 9 Class 11 Sequences and Series - Part 2
Misc 17 - Chapter 9 Class 11 Sequences and Series - Part 3
Misc 17 - Chapter 9 Class 11 Sequences and Series - Part 4

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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

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.