The coefficient of three consecutive terms in the expansion of ${\left(1+a\right)}^n$ are in the ratio $1:7:42$ then n=

Let 3 consecutive terms be ${\mathrm{T}}_{\mathrm{r}-1},{\mathrm{T}}_{\mathrm{r}},{\mathrm{T}}_{\mathrm{r}+1}$ and their coeffi’s are${ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-2}{,}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}{,}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$
$\begin{array}{l}\Rightarrow \frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-2}}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{1}{7}\Rightarrow \frac{r-1}{n-(r-1)+1}=\frac{1}{7}\Rightarrow 7(r-1)=n-r+2\\ \therefore \mathrm{n}-8\mathrm{r}+9=0\text{ And }\frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}=\frac{7}{42}\Rightarrow \frac{r}{n-r+1}=\frac{1}{6}\\ \Rightarrow \mathrm{n}-7\mathrm{r}+1=0\\ 8\left(\mathrm{n}-7\mathrm{r}+1\right)-7\left(\mathrm{n}-8\mathrm{r}+9\right)\Leftarrow n+8-63=0\therefore n=55\end{array}$