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sum of squares of three distinct real numers which are in G.P id S^2. If their sum is alphaS, show that alpha^2 belongs to {1/3,1}u{1,3}.Please elaborate.
sum of squares of three distinct real numers which are in G.P is:S^2 = a^2 + (ar)^2 + (ar^2)^2 = a^2(1 + r^2 + r^4) =a^2(1.(r^2)^3 – 1)/(r^2 – 1)..................(1) Sum is alpha,a(1 + r + r^2) = alphaS ................(2)substituting the eqn. (2) in (1) with value of a and solvng for r that is common ration whic wwould always lie between -1 to 1.The interval for alpha^2 which would always be postive lies in{1/3, 1) U {1,3).
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