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JEE Advanced 2016 Paper 1 · Mathematics · Sequences and Series · medium

Define a1=1 and an+1=1/2(an+3/an) for n≥1.

One or more options correct

Define a1=1a_1=1 and an+1=12(an+3an)a_{n+1}=\dfrac12\left(a_n+\dfrac{3}{a_n}\right) for n≥1n\ge1. Which of the following is/are correct?
  1. A. lim⁡n→∞an=3\displaystyle\lim_{n\to\infty}a_n=\sqrt3
  2. B. The sequence (an)(a_n) is monotonically decreasing
  3. C. a4=2615a_4=\dfrac{26}{15}
  4. D. an<3a_n<\sqrt3 for every n≥2n\ge2
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Answer: A

By AM-GM, an+1=12(an+3an)≥3a_{n+1}=\frac12\left(a_n+\frac3{a_n}\right)\ge\sqrt{3} for every n≥1n\ge1, with equality only if an=3a_n=\sqrt3, which never happens because all terms are rational. Hence an>3a_n>\sqrt3 for n≥2n\ge2, so (D) is false. For n≥2n\ge2 we have an2>3a_n^2>3, so an+1−an=3−an22an<0a_{n+1}-a_n=\frac{3-a_n^2}{2a_n}<0: the sequence decreases from n=2n=2 onwards and, being bounded below by 3\sqrt3, converges to a limit ℓ\ell with ℓ=12(ℓ+3/ℓ)\ell=\frac12(\ell+3/\ell), giving ℓ=3\ell=\sqrt3; so (A) is true. However a1=1<a2=2a_1=1<a_2=2, so the whole sequence is not monotonically decreasing and (B) is false. Computing exactly, a2=2a_2=2, a3=74a_3=\frac74 and a4=12(74+127)=9756a_4=\frac12\left(\frac74+\frac{12}{7}\right)=\frac{97}{56}, not 2615\frac{26}{15}, so (C) is false.
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