The roots of equation $x^5-40x^4+\mathrm{\alpha }{x}^3+\beta x^2+\gamma x+\delta =0$ are real and in GP. If the sum of heir reciprocals is 10, then $\delta$can be 

Given, $x^5-40x^4+\alpha x^3+\beta x^2+\gamma x+\delta =0$

has roots $\frac{a}{r^2},\frac{a}{r},a,ar,a{r}^2$

Now, $\frac{a}{r^2}+\frac{a}{r}+a+ar+a{r}^2=40$   …(i)

and $\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}=10$  ….(ii)

On dividing Eq. (i) by Eq. (ii), we get 

$a^2=4\Rightarrow a=\pm 2$

A.s, the product of roots $=a^5=-\delta$

$\therefore$ $\delta =32\text{ or }-32$