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

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Answer: A. 3654 \begin{aligned} & 	ext { 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) !} 	imes \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} \begin{aligned} & 	ext { Also, } \frac{n}{r !(n-r) !} 	imes \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} From (i) and (ii), we get \begin{aligned} & 6 r-1=5 r+4 \\ & \Rightarrow r=5 \\ & 	ext { So, } n=5(5)+4=29 \\ & 	ext { So, coefficient of } 4{ }^{	ext {th }} 	ext { terms }={ }^n \mathrm{C}_3={ }^{29} \mathrm{C}_3 \\ & =\frac{29 !}{3 ! 26 !}=\frac{29 	imes 28 	imes 27}{3 	imes 2}=3654 \end{aligned}

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

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