If the roots of $x^5-a{x}^4+b{x}^3-c{x}^2+dx-1=0$  are all positive such that their arithmetic mean and geometric are equal then  $a+b+c+d=$

$\text{Let}{x}_1,x_2,x_3,x_4,x_5\text{are roots of}{x}^5-a{x}^4+b{x}^3-c{x}^2 +dx-1=0$
$\text{A.}\mathrm{M}\text{and G.M of}{x}_1,x_2,x_2,x_4,x_5\text{are equal then }$
$\begin{array}{rl} & x_1=x_2=x_3=x_4=x_5=1\\ & \text{Sum of roots}{=}^5{C}_1=5\\ & \Rightarrow a=5\end{array}$
$\text{Sum of roots taken two at a time}{=}^5{C}_2=10$
$\Rightarrow b=10$
$\text{Sum of roots taken three at a time}{=}^5{C}_3=10$
$\Rightarrow c=10$
$\text{Sum of roots four two at a time}{=}^5{C}_4=5$
$\begin{array}{rl} & \Rightarrow d=5\\ & \therefore a+b+c+d=30\end{array}$