if a,b,c,d are in gpthn prove that [a2-b2] ,[b2-c2],[c2-d2]are in ap Share with your friends Share 0 Ashwini Kumar answered this We have,a, b, c, d are in GP.i.e. a, b, c are in GP⇒a2, b2, c2 are in GP Square of a GP is also a GPAgain, b, c, d are in GP.⇒b2, c2, d2 are in GPNow, a2, b2, c2 are in GP and b2, c2, d2 are in GP∴a2-b2, b2-c2, c2-d2 are in GP. Difference of two GPs are also a GP -1 View Full Answer Bhoomika answered this Can you tell me the question is from which book. 0 Umang answered this Is the question correct? I think it should be to prove that the series is in GP 0
