Let $t_p \& t_q$ is the $p^{th} \& q^{th }$terms of the series $-16,8,-4,2,...$

Here $a=-16, r=-\frac{1}{2}$

Then, $t_p=-16{\left(-\frac{1}{2}\right)}^{p-1}$

$t_q=-16{\left(-\frac{1}{2}\right)}^{q-1}$

The roots of the given quadratic equation $4x^2-9x+5=0$ is $x=1,\frac{5}{4}.$

$AM=\frac{5}{4}, GM=1 \left(\because AM\ge GM\right)$

Now, given $\frac{5}{4}=\frac{t_p+t_q}{2},1=t_p{t}_q$

So, $1=256{\left(-\frac{1}{2}\right)}^{p+q-2}\Rightarrow 2^{p+q-2}={\left(-1\right)}^{p+q-2}{2}^8$

$\Rightarrow 2^{p+q-10}={\left(-1\right)}^{p+q-2}$

It is equal when $2^{p+q-10}=1={\left(-1\right)}^{p+q-2}$

Hence, $p+q-10=0\Rightarrow p+q=10$