Let An = (3/4) - (3/4)² + (3/4)³ - ... + (-1)^(n+1)(3/4)^n and Bn = 1 - An. Then, the least odd natural number p, so that Bn > An for all n ≥ p, is

Formula: Leta,ar,ar2+ar3+...+arn𕒵benterms of a a GP. Then its sum is given by,S=a(1−rn)1−rGiven,An=(34)−(34)2+(34)3−+(𕒵)n𕒵(34)nIt is a Geometric Progression (GP) witha=34,r=󔼪and number of terms=nTherefore,An=34×(1−(󔼪)n)1−(󔼪)⇒An=34×(1−(󔼪)n)74⇒An=37[1−(󔼪)n]Also given,Bn=1−AnTo find: The least odd natural numberp, such thatBn>AnNow,1−An>An⇒1>2×An⇒An<12Substituting the value ofAnin the above equation, we get37×[1−(󔼪)n]<12⇒1−(󔼪)n<76⇒1󔽔<(󔼪)n⇒𕒵6<(󔼪)nSincenis odd, then(󔼪)n=(𕒵)×34nTherefore,𕒵6<(𕒵)×(34)nMultiplying the entire inequality by𕒵, we get16>(34)nNow, Applying log to the base34log3416<34⇒6.228<nTherefore,nshould be7.