The set of all values of ‘a’ for which the roots of the equation $(a+1)x^2-3ax+4a=0(a\ne -1)\text{ are real and greater than }1\text{ is }$

$\begin{array}{l}(a+1)x^2-3ax+4a=0\\ D=9a^2-16a(a+1)\ge 0\\ \Rightarrow -7a^2-16a\ge 0 \Rightarrow a\left(7a+16\right)\le 0\\ \Rightarrow a\in \left[\frac{-16}{7},0\right].......\left(1\right)\\ x_1>1,x_2>1\\ \therefore (a+1)f\left(1\right)>0\text{ and }\frac{3a}{2(a+1)}>1\\ \Rightarrow (a+1)(2a+1)>0\end{array}$

$\begin{array}{l}\text{ a<-1 or a >}\frac{\text{-1}}{2}\text{ and }\frac{a-2}{a+1}>0\dots \dots ...\left(2\right)\\ \\ \text{from }\left(1\right)\text{,}\left(2\right)\\ \therefore a\in \left[-\frac{16}{7},-1\right)\end{array}$