If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }, \mathrm{\delta }$ are the roots of ${\mathrm{x}}^4–10{\mathrm{x}}^3+35{\mathrm{x}}^2–50\mathrm{x}+24=0$ which are in A.P. where $\mathrm{\alpha }=1$ then common difference of that A.P is $(\mathrm{\alpha }<\mathrm{\beta }<\mathrm{\gamma }<\mathrm{\delta })$ 

$\begin{array}{l}Let\delta =a+3d,\gamma =a+d,\beta =a-d,\alpha =a-3darerootsof{x}^4-10x^3+35x^2-50x+24=0\\ \therefore \alpha +\beta +\gamma +\delta =10\\ \Rightarrow 4a=10\Rightarrow a=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ andgiven\alpha =1\\ \Rightarrow a-3d=1\\ \Rightarrow \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-3d=1\\ \Rightarrow 3d=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1\Rightarrow \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \Rightarrow d=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \therefore Commondifference=2d\Rightarrow 1\end{array}$