A) \[\frac{25840}{9}\]
B) \[\frac{24840}{9}\]
C) \[\frac{26840}{9}\]
D) None of these
Correct Answer: A
Solution :
Let (r + 1)th term be the greatest term. Then \[{{T}_{r+1}}=\sqrt{3}.{{\,}^{20}}{{C}_{r}}{{\left( \frac{1}{\sqrt{3}} \right)}^{r}}\]and \[{{T}_{r}}=\sqrt{3}.\,{{\,}^{20}}{{C}_{r-1}}{{\left( \frac{1}{\sqrt{3}} \right)}^{r-1}}\] Now \[\frac{{{T}_{r+1}}}{{{T}_{r}}}=\frac{20-r+1}{r}\left( \frac{1}{\sqrt{3}} \right)\] \[\therefore \,\,\,\,{{T}_{r+1}}\ge {{T}_{r}}\Rightarrow 20-r+1\ge \sqrt{3}r\] \[\Rightarrow 21\ge r(\sqrt{3}+1)\,\,\Rightarrow r\le \frac{21}{\sqrt{3}+1}\]\[\Rightarrow \,\,r\le 7.686\Rightarrow r=7\] Hence the greatest term is \[{{T}_{8}}=\sqrt{3}{{\,}^{20}}{{C}_{7}}{{\left( \frac{1}{\sqrt{3}} \right)}^{7}}=\frac{25840}{9}\]
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