The largest coefficient in the expansion of $(4+3\mathrm{x}{)}^{25}$ is

We have $(4+3\mathrm{x}{)}^{25}=4^{25}{\left(1+\frac{3}{4}\mathrm{x}\right)}^{25}$

Let $(\mathrm{r}+1{)}^{\mathrm{th}}$ is term is the term having greatest coefficient

$\Rightarrow$coefficient of ${\mathrm{T}}_{\mathrm{r}+1}\ge$ coefficient of ${\mathrm{T}}_{\mathrm{r}}$

$\begin{array}{l}\Rightarrow 4^{25}\left\{{ }^{25}{\mathrm{C}}_{\mathrm{r}}{\left(\frac{3}{4}\right)}^{\mathrm{r}}\right\}\ge 4^{25}\left\{{ }^{25}{\mathrm{C}}_{\mathrm{r}-1}{\left(\frac{3}{4}\right)}^{\mathrm{r}-1}\right\}\\ \Rightarrow \frac{{ }^{25}{\mathrm{C}}_{\mathrm{r}}}{{ }^{25}{\mathrm{C}}_{\mathrm{r}-1}}\left(\frac{3}{4}\right)\ge 1\Rightarrow \frac{25-(\mathrm{r}-1)}{\mathrm{r}}\frac{3}{4}\ge 1\\ \Rightarrow 75-3\mathrm{r}+3\ge 4\mathrm{r}\therefore \mathrm{r}\le \frac{78}{7}\le 11.142\end{array}$

But, r is an integer, hence $\mathrm{r}=11$.