If a_ka_k-1 + a_k-1​a_k-2​ = 2a​_k a_k-2  , k>=3 and a₁=1 , here S_p= sigma (k = 1 to k = p) 1/a_k and given that S_2p/S_p does not depend on p then 1/a₂₀₁₆ is = ?

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}+{\mathrm{a}}_{\mathrm{k}-1} {\mathrm{a}}_{\mathrm{k}-2}=2{\mathrm{a}}_{\mathrm{k}} {\mathrm{a}}_{\mathrm{k}-2} \\ \mathrm{Put} \mathrm{k}=3 \\ \Rightarrow {\mathrm{a}}_3{\mathrm{a}}_2+{\mathrm{a}}_2 {\mathrm{a}}_1=2{\mathrm{a}}_3 {\mathrm{a}}_1 \\ \Rightarrow {\mathrm{a}}_3{\mathrm{a}}_2+{\mathrm{a}}_2=2{\mathrm{a}}_3 \left[\mathrm{As} {\mathrm{a}}_1=1 \right] \\ \Rightarrow {\mathrm{a}}_3=\frac{{\mathrm{a}}_2}{2-{\mathrm{a}}_2} ;\left(\mathrm{i}\right) \\ \mathrm{Similarly} \mathrm{after} \mathrm{putting} \mathrm{k}=4, \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}_4{\mathrm{a}}_3+{\mathrm{a}}_3 {\mathrm{a}}_2=2{\mathrm{a}}_4 {\mathrm{a}}_2 \\ \Rightarrow {\mathrm{a}}_4=\frac{{\mathrm{a}}_3 {\mathrm{a}}_2}{2{\mathrm{a}}_2-{\mathrm{a}}_3} \\ \Rightarrow {\mathrm{a}}_4=\frac{\left(\frac{{\mathrm{a}}_2}{2-{\mathrm{a}}_2}\right) {\mathrm{a}}_2}{2{\mathrm{a}}_2-\left(\frac{{\mathrm{a}}_2}{2-{\mathrm{a}}_2}\right)}=\frac{{{\mathrm{a}}_2}^2}{3{\mathrm{a}}_2-2{{\mathrm{a}}_2}^2}=\frac{{\mathrm{a}}_2}{3-2{\mathrm{a}}_2} ;\left(\mathrm{ii}\right) \\ \mathrm{Also} \mathrm{according} \mathrm{to} \mathrm{ques}: \\ \frac{{\mathrm{S}}_{2\mathrm{p}}}{{\mathrm{S}}_{\mathrm{p}}} \mathrm{is} \mathrm{independent} \mathrm{of} \mathrm{p} \\ \Rightarrow \frac{{\mathrm{S}}_2}{{\mathrm{S}}_1}=\frac{{\mathrm{S}}_4}{{\mathrm{S}}_2} \\ \Rightarrow \frac{1+{\displaystyle \frac{1}{{\mathrm{a}}_2}}}{1}=\frac{1+{\displaystyle \frac{1}{{\mathrm{a}}_2}}+{\displaystyle \frac{2-{\mathrm{a}}_2}{{\mathrm{a}}_2}}+{\displaystyle \frac{3-2{\mathrm{a}}_2}{{\mathrm{a}}_2}}}{1+{\displaystyle \frac{1}{{\mathrm{a}}_2}}} \\ \Rightarrow {\left(1+\frac{1}{{\mathrm{a}}_2}\right)}^2=\left(1+\frac{1}{{\mathrm{a}}_2}\right)+\frac{2-{\mathrm{a}}_2}{{\mathrm{a}}_2}+\frac{3-2{\mathrm{a}}_2}{{\mathrm{a}}_2} \\ \Rightarrow {\left(1+\frac{1}{{\mathrm{a}}_2}\right)}^2-\left(1+\frac{1}{{\mathrm{a}}_2}\right)=\frac{2-{\mathrm{a}}_2+3-2{\mathrm{a}}_2}{{\mathrm{a}}_2} \\ \Rightarrow \left(1+\frac{1}{{\mathrm{a}}_2}\right)\left[1+\frac{1}{{\mathrm{a}}_2}-1\right]=\frac{5-3{\mathrm{a}}_2}{{\mathrm{a}}_2} \\ \Rightarrow \left(1+\frac{1}{{\mathrm{a}}_2}\right)\left[\frac{1}{{\mathrm{a}}_2}\right]=\frac{5-3{\mathrm{a}}_2}{{\mathrm{a}}_2} \\ \Rightarrow \frac{{\mathrm{a}}_2+1}{{\mathrm{a}}_2}=5-3{\mathrm{a}}_2 \\ \Rightarrow 3{{\mathrm{a}}_2}^2-4{\mathrm{a}}_2+1=0 \\ \Rightarrow 3{{\mathrm{a}}_2}^2-3{\mathrm{a}}_2-{\mathrm{a}}_2+1=0 \\ \Rightarrow 3{\mathrm{a}}_2\left({\mathrm{a}}_2-1\right)-1\left({\mathrm{a}}_2-1\right)=0 \\ \Rightarrow \left({\mathrm{a}}_2-1\right)\left(3{\mathrm{a}}_2-1\right)=0 \\ \Rightarrow {\mathrm{a}}_2=1 \mathrm{or} {\mathrm{a}}_2=\frac{1}{3} \\ \mathrm{But} \mathrm{if} {\mathrm{a}}_2=1, \mathrm{then} \mathrm{series} \mathrm{will} \mathrm{become} 1,1,1,1,1,1,1, \\ \mathrm{Hence} \mathrm{we} \mathrm{take} {\mathrm{a}}_2=\frac{1}{3} \\ \mathrm{Putting} \mathrm{value} \mathrm{of} {\mathrm{a}}_2 \mathrm{in} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}_3=\frac{{\displaystyle \frac{1}{3}}}{2-{\displaystyle \frac{1}{3}}}=\frac{1}{5} \\ {\mathrm{a}}_4=\frac{{\displaystyle \frac{1}{3}}}{3-{\displaystyle \frac{2}{3}}}=\frac{1}{7} \\ \mathrm{Hence} \mathrm{our} \mathrm{series} \mathrm{is} 1,\frac{1}{3},\frac{1}{5},\frac{1}{7},..... \\ \mathrm{As} \mathrm{denominator}\text{'}\mathrm{s} \mathrm{term} \mathrm{are} \mathrm{in} \mathrm{A}.\mathrm{P}. \mathrm{wil} \mathrm{first} \mathrm{term}=1 \mathrm{and} \mathrm{common} \mathrm{difference}=2 \\ \mathrm{Hence} \mathrm{we} \mathrm{get}: \\ {\mathrm{a}}_{2016}=\frac{1}{1+\left(2016-1\right)2}=\frac{1}{4031} \\ \mathrm{Hence} \\ \frac{\mathbf{1}}{\mathbf{ }{\mathbf{a}}_{\mathbf{2016}}}\mathbf{=}\mathbf{4031}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{Note}: \mathrm{In} \mathrm{case} \mathrm{series} \mathrm{is} \mathrm{taken} \mathrm{as} 1,1,1,1,1,1,..., \mathrm{then} \frac{1}{ {\mathrm{a}}_{2016}}=1 \end{gathered}$$

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