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

$$ \begin{aligned} & \text { Given: }{ }^n \mathrm{C}_{r-1}:{ }^n \mathrm{C}_r:{ }^n \mathrm{C}_{r+1} \\\\ & =1: 5: 20 \\\\ & \Rightarrow \frac{n !}{(r-1) !(n-r+1) !} \times \frac{r !(n-r) !}{n !}=\frac{1}{5} \\\\ & \Rightarrow \frac{r}{(n-r+1)}=\frac{1}{5} \\\\ & \Rightarrow 5 r=n-r+1 \\\\ & \Rightarrow n=6 r-1 ........(i) \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Also, } \frac{n}{r !(n-r) !} \times \frac{(r+1) !(n-r-1) !}{n !}=\frac{5}{20}=\frac{1}{20} \\\\ & \Rightarrow \frac{(r+1)}{(n-r)}=\frac{1}{4} \\\\ & \Rightarrow 4 r+4=n-r \\\\ & \Rightarrow n=5 r+4 ..........(ii) \end{aligned} $$ <br/><br/>From (i) and (ii), we get <br/><br/>$$ \begin{aligned} & 6 r-1=5 r+4 \\\\ & \Rightarrow r=5 \\\\ & \text { So, } n=5(5)+4=29 \\\\ & \text { So, coefficient of } 4{ }^{\text {th }} \text { terms }={ }^n \mathrm{C}_3={ }^{29} \mathrm{C}_3 \\\\ & =\frac{29 !}{3 ! 26 !}=\frac{29 \times 28 \times 27}{3 \times 2}=3654 \end{aligned} $$