Explanation

(c)
$T_{r + 1} = \sqrt{3} \bullet \ ^{20}C_{r}\left( \frac{1}{\sqrt{3}} \right)^{r}$
$and\ T_{r} = \sqrt{3} \bullet \ ^{20}C_{r - 1}\left( \frac{1}{\sqrt{3}} \right)^{r - 1}$
$Now,\frac{T_{r + 1}}{T_{r}} = \frac{20 - r + 1}{r}\left( \frac{1}{\sqrt{3}} \right)$
$Since,\ T_{r + 1} \geq T_{r} \Longrightarrow 20 - r + 1 \geq \sqrt{3}r$
$\Longrightarrow r \leq \frac{21}{\sqrt{3} + 1} = \frac{21}{2 \bullet 73} \Longrightarrow r \leq 7.692$
$\Longrightarrow r = 7$
$\therefore$ The greatest term is
$\ \ \ \ \ \ \ \ \ \ \ T_{3} = \sqrt{3} \bullet \ ^{20}C_{7}\left( \frac{1}{\sqrt{3}} \right)^{7} = \frac{25840}{9\ }\$