Greatest term in the binomial expansion of ${(a + 2x)}^9$ when $a = 1 \& x = \frac{1}{3}$ is :

$$\begin{gathered} \frac{T_{r+1}}{T_r}=\frac{{}^9{C}_r{a}^{9-r}{\left(2x\right)}^r}{C_r{a}^{9-r+1}{\left(2x\right)}^{r-1}}=\frac{9-r+1}{r}\frac{2x}{a}=\frac{9-r+1}{r}\frac{2}{3}=\frac{10-r}{r}\frac{2}{3} \\ \therefore T_{r+1}>T_r if \frac{10-r}{r}\frac{2}{3}>1 i.e. 20>5r i.e. r<4 \end{gathered}$$

Hence for values of $\text{'}r\text{'}$ upto $3, T_{r + 1}< T_r$

when $r = 4, T_{r+1} = T_r i.e. T_4= T_5$& both of them are the greatest terms