A) 3
B) -3
C) 6
D) -6
Correct Answer: A
Solution :
\[{{x}^{2}}-6x-2=0\] has roots \[\alpha ,\beta \] ?..(1) \[{{a}_{n}}={{\alpha }^{n}}-{{\beta }^{n}}\forall x\ge 1\] RAO IIT ACADEMY / JEE - Now \[\frac{{{a}_{10}}-2.{{a}_{8}}}{2.{{a}_{9}}}\] \[=\frac{\left( {{\alpha }^{10}}-{{\beta }^{10}} \right)-2\left( {{\alpha }^{8}}-{{\beta }^{8}} \right)}{2.\left( {{\alpha }^{9}}-{{\beta }^{9}} \right)}\] Using ?2 as \[\alpha .\beta \] \[\Rightarrow \] \[\frac{\left( {{\alpha }^{10}}-{{\beta }^{10}} \right)+\alpha .\beta \left( {{\alpha }^{8}}-{{\beta }^{8}} \right)}{2.\left( {{\alpha }^{9}}-{{\beta }^{9}} \right)}\] \[\Rightarrow \]\[\frac{{{\alpha }^{9}}\left( \alpha +\beta \right)-{{\beta }^{9}}\left( \beta +\alpha \right)}{2.\left( {{\alpha }^{9}}-{{\beta }^{9}} \right)}\] \[\Rightarrow \]\[\frac{\left( {{\alpha }^{9}}-{{\beta }^{9}} \right).\left( \alpha +\beta \right)}{2.\left( {{\alpha }^{9}}-{{\beta }^{9}} \right)}=\frac{\alpha +\beta }{2}\Rightarrow \frac{6}{2}=3\]
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