1. Identify the sequence

The sequence is $-16,\ 8,\ -4,\ 2,\ \ldots$

This is a geometric progression with:

• first term $a=-16$
• common ratio $r=\frac{8}{-16}=-\frac12$

So the $n$th term is $T_n=a r^{n-1}=-16\left(-\frac12\right)^{n-1}$

Thus, $T_p=-16\left(-\frac12\right)^{p-1}, \qquad T_q=-16\left(-\frac12\right)^{q-1}$

2. Use the condition on arithmetic mean and geometric mean

Let the arithmetic mean and geometric mean of $T_p$ and $T_q$ be the two roots of $4x^2-9x+5=0$

Factorizing, $4x^2-9x+5=(4x-5)(x-1)$

So the roots are $1,\ \frac54$

Hence the arithmetic mean (AM) and geometric mean (GM) are $1$ and $\frac54$ in some order.

Now for two numbers, AM $\ge$ GM (when both are non-negative). Since both roots are positive and $\frac54>1$, we must have $\text{AM}=\frac54, \qquad \text{GM}=1$

So, $\frac{T_p+T_q}{2}=\frac54 \quad \Rightarrow \quad T_p+T_q=\frac52$

and $\sqrt{T_pT_q}=1 \quad \Rightarrow \quad T_pT_q=1$

3. Find the two terms using sum and product

If two numbers have sum $\frac52$ and product $1$, then they satisfy $t^2-\frac52 t+1=0$

Multiply by $2$: $2t^2-5t+2=0$

Factorize: $2t^2-5t+2=(2t-1)(t-2)=0$

Thus the two terms are $T_p=2, \qquad T_q=\frac12$

(in any order).

4. Locate these terms in the GP

Sequence terms are: $T_1=-16,\ T_2=8,\ T_3=-4,\ T_4=2,\ T_5=-1,\ T_6=\frac12,\ldots$

Therefore, $2=T_4, \qquad \frac12=T_6$

So, $p=4, \qquad q=6$

Hence, $p+q=4+6=10$

5. Comparison with stored answer

Derived answer is $\boxed{10}$

This matches the stored correct answer.