If the three successive coefficients in the binomial expansion of ${(1+x)}^n$ are 28, 56 and 70 respectively, then $\text{'}n\text{'}$ is

$\begin{array}{l}{T}_r={}^n{\text{C}}_{r-1}=28\cdots \cdots \cdots \left(1\right)\\ T_{r+1}={}^n{\text{C}}_r=56\cdots \cdots \cdots \left(2\right)\\ T_{r+2}={}^n{\text{C}}_{r+1}=70\cdots \cdots \cdots \left(3\right)\end{array}$

$\begin{array}{l}\frac{\left(2\right)}{\left(1\right)}\Rightarrow \frac{{}^n{\text{C}}_r}{{}^n{\text{C}}_{r-1}}=\frac{56}{28}\\ \frac{n-r+1}{r}=2\\ n-r+1=2r\\ 3r=n+1\\ r=\frac{n+1}{3}\cdots \cdots \cdots \left(4\right)\end{array}$

$\begin{array}{l}\frac{\left(3\right)}{\left(2\right)}\Rightarrow \frac{{}^n{\text{C}}_{r+1}}{{}^n{\text{C}}_r}=\frac{70}{56}\\ \frac{n-r}{r+1}=\frac{5}{4}\\ 4n-4r=5r+5\\ 9r=4n-5\end{array}$

$\begin{array}{l}9\left(\frac{n+1}{3}\right)=4n-5\left[\text{from }\left(\right)\right]\\ 3n+3=4n-5\\ n=8\end{array}$