Given equation is $x^5-a x^4+b x-c x+d x-1=0$.
Here, geometrical mean and arithmetic mean are equal then all the roots are equal.
Let $\alpha$ is the root of the equation.
$\frac{\alpha+\alpha+\alpha+\alpha+\alpha}{5}=\left(\alpha^5\right)^{1 / 5}$
Here, sum of all roots is $a$.
$\begin{gathered}
\frac{a}{5}=\alpha \\
a=5 \alpha
\end{gathered}$
$(\alpha \alpha \alpha \alpha \alpha)^{1 / 5}=\frac{-e}{a}=-(-1)=1$
$\begin{aligned}
& \alpha^{5 \times \frac{1}{5}}=1 \\
& \alpha=1
\end{aligned}$
From (i), $a=5$.
Now,
$\begin{aligned}
& \alpha \alpha+\alpha \alpha+\alpha \alpha+\ldots . .+\alpha \alpha(10 \text { times })=b \\
& 10 \alpha^2=b \\
& b=10 \\
& \alpha \alpha \alpha+\alpha \alpha \alpha+\ldots . .+10 \text { times }=c \\
& 10 \alpha^3=c \\
& c=10 \\
& \alpha \alpha \alpha \alpha+\alpha \alpha \alpha \alpha+\ldots \ldots \ldots+5 \text { times }=d \\
& 5 \alpha^4=d \\
& d=5
\end{aligned}$
Therefore, $a+d+c+b=5+5+10+10=30$.