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Vijay posted an Question
October 26, 2021 • 06:13 am 30 points
  • IIT JAM
  • Mathematics (MA)

Vmcq11. please send the solution to the problem given in the attachment.

2 Answer(s) Answer Now
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    Arpan pal best-answer

    option (d) is correct. since Xn+1 =(Xn+Yn)/2 and Yn+1=√(Xn.Yn) Then applying A.M≥G.M we have Xn+1=(Xn+Yn)/2 ≥√(Xn.Yn)=Yn+1 for all n≥2 i.e.,Xn≥Yn for all n≥1 Taking limit both sides, we have limXn≥limYn .

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    C is correct option acc.to the book

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    pls explain the last 3rd step

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    Xn+1≥Yn+1.It is true for all n≥2.If you put n+1=n,then Xn≥Yn and this true for all n+1≥2 i.e.,n≥1

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