If the arithmetic mean and the geometric mean of the $$p^{th}$$ and $$q^{th}$$ terms of the sequence $$-16, 8, -4, 2, \ldots$$ satisfy the equation $$4x^2 - 9x + 5 = 0$$, then $$p + q$$ is equal to ______.

The sequence $$-16, 8, -4, 2, \ldots$$ is a geometric progression with first term $$a_1 = -16$$ and common ratio $$r = -\frac{1}{2}$$. The general term is $$a_n = -16 \cdot \left(-\frac{1}{2}\right)^{n-1} = (-1)^n \cdot 2^{5-n}$$.

The equation $$4x^2 - 9x + 5 = 0$$ factors as $$(4x - 5)(x - 1) = 0$$, giving roots $$x = \frac{5}{4}$$ and $$x = 1$$. These are the AM and GM of $$a_p$$ and $$a_q$$.

For the GM to be real, we need $$a_p \cdot a_q > 0$$. We have $$a_p \cdot a_q = (-16)^2 \cdot \left(-\frac{1}{2}\right)^{p+q-2} = 256 \cdot (-1)^{p+q-2} \cdot \left(\frac{1}{2}\right)^{p+q-2} = 256 \cdot (-1)^{p+q} \cdot \left(\frac{1}{2}\right)^{p+q-2}$$. This is positive when $$p + q$$ is even.

The geometric mean is $$\text{GM} = \sqrt{a_p \cdot a_q} = \sqrt{256 \cdot \left(\frac{1}{2}\right)^{p+q-2}} = 16 \cdot \left(\frac{1}{2}\right)^{(p+q-2)/2} = 16 \cdot 2^{-(p+q-2)/2} = 2^{4-(p+q-2)/2} = 2^{5-p/2-q/2}$$.

If $$\text{GM} = 1 = 2^0$$, then $$5 - \frac{p+q}{2} = 0$$, giving $$p + q = 10$$.

Now we verify the AM. With $$p + q = 10$$, the AM equals $$\frac{a_p + a_q}{2}$$. Since the terms have equal product signs and GM = 1, both $$a_p$$ and $$a_q$$ must be positive (as their product is positive and GM is positive). The AM-GM inequality gives $$\text{AM} \geq \text{GM}$$, so $$\text{AM} \geq 1$$. Since $$\frac{5}{4} > 1$$, we assign $$\text{AM} = \frac{5}{4}$$ and $$\text{GM} = 1$$. This is consistent because AM $$\cdot$$ GM $$= \frac{5}{4}$$ equals the product of roots of the quadratic, and AM $$+$$ GM $$= \frac{9}{4}$$ equals the sum of roots.

For example, $$p = 4, q = 6$$: $$a_4 = 2, a_6 = \frac{1}{2}$$. Then AM $$= \frac{2 + 1/2}{2} = \frac{5}{4}$$ and GM $$= \sqrt{2 \cdot \frac{1}{2}} = 1$$. Both satisfy $$4x^2 - 9x + 5 = 0$$.

Therefore, $$p + q = 10$$.