The coefficients of three consecutive terms in the expansion of $(1+x)^n$ are in the ratio $1:7:42$, then the value of $n$ is

Let coefficients be $ ^nC_{r-2},\ ^nC_{r-1},\ ^nC_r $ Given ratio: $ \frac{^nC_{r-2}}{^nC_{r-1}}=\frac{1}{7} $ $ \Rightarrow \frac{r-1}{n-r+2}=\frac{1}{7} $ $ \Rightarrow 7r-7=n-r+2 $ $ \Rightarrow n-8r+9=0 \quad ...(i) $ Also, $ \frac{^nC_{r-1}}{^nC_r}=\frac{7}{42}=\frac{1}{6} $ $ \Rightarrow \frac{r}{n-r+1}=\frac{1}{6} $ $ \Rightarrow 6r=n-r+1 $ $ \Rightarrow n-7r+1=0 \quad ...(ii) $ Solving (i) and (ii), $ r=8,\ n=55 $