if the coefficients of three consecutive terms in the expansion of 1 + x n are the ratio 1   :   5…

Given,

${}^{n}C_{r-1}:{}^{n}C_r:{}^{n}C_{r+1}=1:5:20$

Now using the formula $\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}$ we get,

.$\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}=\frac{5}{1}$

$\Rightarrow n-6r+1=0 .........\left(1\right)$

And $\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}=\frac{20}{5}$

$\Rightarrow n-5r-4=0 ..........\left(2\right)$

Solving above equations we get,

$n=29 \& r=5$

So, the coefficient of forth term will be ${}^{n}C_3={}^{29}C_3=\frac{29\times 28\times 27}{3\times 2\times 1}=3654$