If the coefficients of three consecutive terms in the expansion of ${(1+\mathrm{x})}^{\mathrm{n}}$ are in the ratio 1 : 5 : 20, then the coefficient of the fourth term is 

${ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}{:}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{:}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}=1:5:20$
$\begin{array}{l}\frac{{ }^n{C}_{r-1}}{1}=\frac{{ }^n{C}_r}{5}=\frac{{ }^n{C}_{r+1}}{20}\\ \because \frac{{ }^n{C}_r}{{ }^n{C}_{r-1}}=5\\ \Rightarrow \frac{n-r+1}{r}=5\Rightarrow n=6r-1\dots \text{ (i) }\\ \because \frac{{ }^n{C}_{r+1}}{{ }^n{C}_r}=4\Rightarrow \frac{n-r}{r+1}=4\Rightarrow n=5r+4\end{array}$
$\text{ from (i) and (ii), }r=5,n=29$
$\Rightarrow \text{ Coefficient of fourth term }{=}^{29}{C}_3=3654$