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Question Number 16110 by vpawarksp@gmail.com last updated on 18/Jun/17
let a_1 >a_2 >0 and a_(n+1) =(√(a_n a_(n−1   ) ))  where n is greater than equal to 2   Then  The sequence {a_(2n) } is   (1) monotonic increasing  (2)monotonic decreasing  (3)non monotonic  (4)unbounded
$$\mathrm{let}\:\mathrm{a}_{\mathrm{1}} >\mathrm{a}_{\mathrm{2}} >\mathrm{0}\:\mathrm{and}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\sqrt{\mathrm{a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}−\mathrm{1}\:\:\:} } \\ $$$$\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2}\: \\ $$$$\mathrm{Then} \\ $$$$\mathrm{The}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\mathrm{2n}} \right\}\:\mathrm{is}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{monotonic}\:\mathrm{increasing} \\ $$$$\left(\mathrm{2}\right)\mathrm{monotonic}\:\mathrm{decreasing} \\ $$$$\left(\mathrm{3}\right)\mathrm{non}\:\mathrm{monotonic} \\ $$$$\left(\mathrm{4}\right)\mathrm{unbounded} \\ $$$$ \\ $$
Commented by RasheedSoomro last updated on 18/Jun/17
let a_1 >a_2 >0 and a_(n+1) =(√(a_n a_(n−1   ) ))  a_n =(((a_(n+1) )^2 )/a_(n−1) )...............(i)  n→n−1⇒a_n =(√(a_(n−1) a_(n−2) )).........(ii)  (i) & (ii)  (((a_(n+1) )^2 )/a_(n−1) )=(√(a_(n−1) a_(n−2) ))
$$\mathrm{let}\:\mathrm{a}_{\mathrm{1}} >\mathrm{a}_{\mathrm{2}} >\mathrm{0}\:\mathrm{and}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\sqrt{\mathrm{a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}−\mathrm{1}\:\:\:} } \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{a}_{\mathrm{n}−\mathrm{1}} }……………\left(\mathrm{i}\right) \\ $$$$\mathrm{n}\rightarrow\mathrm{n}−\mathrm{1}\Rightarrow\mathrm{a}_{\mathrm{n}} =\sqrt{\mathrm{a}_{\mathrm{n}−\mathrm{1}} \mathrm{a}_{\mathrm{n}−\mathrm{2}} }………\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{i}\right)\:\&\:\left(\mathrm{ii}\right) \\ $$$$\frac{\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} \right)^{\mathrm{2}} }{\mathrm{a}_{\mathrm{n}−\mathrm{1}} }=\sqrt{\mathrm{a}_{\mathrm{n}−\mathrm{1}} \mathrm{a}_{\mathrm{n}−\mathrm{2}} } \\ $$$$ \\ $$

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