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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)/(π‘Ž^𝑛+𝑏^𝑛 ) = (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Need to show (𝑏^𝑛+𝑐^𝑛)/(π‘Ž^𝑛+𝑏^𝑛 ) = (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Taking L.H.S (𝑏^𝑛+𝑐^𝑛)/(π‘Ž^𝑛+𝑏^𝑛 ) Putting b = ar , c = ar2 = (γ€–(π‘Žπ‘Ÿ)γ€—^𝑛+γ€–(γ€–π‘Žπ‘Ÿγ€—^2)γ€—^𝑛)/(π‘Ž^𝑛+γ€–(π‘Žπ‘Ÿ)γ€—^𝑛 ) = (π‘Ž^𝑛 π‘Ÿ^𝑛 + π‘Ž^𝑛.π‘Ÿ^2𝑛)/(π‘Ž^𝑛 + π‘Ž^𝑛 π‘Ÿ^𝑛 ) = (π‘Ž^𝑛 π‘Ÿ^𝑛 (1 +π‘Ÿ^𝑛))/(π‘Ž^𝑛 (1 +π‘Ÿ^𝑛)) = rn Taking R.H.S (𝑐^𝑛+𝑑^𝑛)/(𝑏^𝑛+𝑐^𝑛 ) Putting c = ar2, d = ar3, b = ar = (γ€–(γ€–π‘Žπ‘Ÿγ€—^2)γ€—^𝑛 + γ€–(γ€–π‘Žπ‘Ÿγ€—^3)γ€—^𝑛)/(γ€–(π‘Žπ‘Ÿ)γ€—^𝑛 +γ€– (γ€–π‘Žπ‘Ÿγ€—^2)γ€—^𝑛 ) = (π‘Ž^𝑛 π‘Ÿ^2𝑛+ π‘Ž^𝑛 .π‘Ÿ^2𝑛)/(π‘Ž^𝑛 π‘Ÿ^𝑛+γ€– π‘Žγ€—^𝑛 π‘Ÿ^2𝑛 ) = (π‘Ž^𝑛 (π‘Ÿ^2𝑛 + π‘Ÿ^3𝑛))/(π‘Ž^𝑛 (π‘Ÿ^𝑛 + π‘Ÿ^2𝑛)) = ( (π‘Ÿ^2𝑛 + π‘Ÿ^3𝑛))/((π‘Ÿ^𝑛 + π‘Ÿ^2𝑛)) = (π‘Ÿ^2𝑛 (1 + π‘Ÿ))/(π‘Ÿ^𝑛 (1 + π‘Ÿ)) = π‘Ÿ^2𝑛/π‘Ÿ^𝑛 = (π‘Ÿ^𝑛.π‘Ÿ^𝑛)/π‘Ÿ^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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.