anthropos Posted June 16, 2007 Share Posted June 16, 2007 The first two terms of a geometric progression are x and y where -x < y < x (<--what does this mean?) If the sum of the first n terms is equal to the sum to infinity of the remaining termsm prove that x^n = 2y^n. So I did this. T1 = x = a T2= y r= y/x Tn+1 = y^n ... x(1-(y/x)^n) = y^n (after substituting a and r values into G.P formulae) But i don get the answer! what's the prob? Link to comment Share on other sites More sharing options...
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