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Misc 17 You are here

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Last updated at May 29, 2018 by Teachoo

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, β¦. 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 (π^π+π^π)/(π^π+π^π ) = (π^π+π^π)/(π^π+π^π ) Need to show (π^π+π^π)/(π^π+π^π ) = (π^π+π^π)/(π^π+π^π ) Taking L.H.S (π^π+π^π)/(π^π+π^π ) Putting b = ar , c = ar2 = (γ(ππ)γ^π+γ(γππγ^2)γ^π)/(π^π+γ(ππ)γ^π ) = (π^π π^π + π^π.π^2π)/(π^π + π^π π^π ) = (π^π π^π (1 +π^π))/(π^π (1 +π^π)) = rn Taking R.H.S (π^π+π^π)/(π^π+π^π ) Putting c = ar2, d = ar3, b = ar = (γ(γππγ^2)γ^π + γ(γππγ^3)γ^π)/(γ(ππ)γ^π +γ (γππγ^2)γ^π ) = (π^π π^2π+ π^π .π^2π)/(π^π π^π+γ πγ^π π^2π ) = (π^π (π^2π + π^3π))/(π^π (π^π + π^2π)) = ( (π^2π + π^3π))/((π^π + π^2π)) = (π^2π (1 + π))/(π^π (1 + π)) = π^2π/π^π = (π^π.π^π)/π^π = rn = rn = L.H.S Thus L.H.S = R.H.S Hence (an + bn), (bn + cn) & (cn + bn) are in GP Hence proved