Let the arithmetic mean of $$\dfrac{1}{a}$$ and $$\dfrac{1}{b}$$ be $$\dfrac{5}{16}$$, $$\text{a > 2}$$. If $$\alpha$$ is such that $$ a,\alpha,b $$ are in A.P., then the equation $$\alpha x^{2}-ax+2(\alpha-2b)=0$$ has:

Find relations:

$$\frac{1/a + 1/b}{2} = \frac{5}{16} \implies \frac{a+b}{2ab} = \frac{5}{16} \implies \frac{a+b}{ab} = \frac{5}{8}$$.

Since $$a, \alpha, b$$ are in A.P., $$\alpha = \frac{a+b}{2}$$.

Substitute $$\alpha$$: From the first relation, $$2\alpha / ab = 5/8 \implies ab = \frac{16\alpha}{5}$$.

 Simplify the equation: $$\alpha x^2 - ax + 2(\alpha - 2b) = 0$$.

Let $$g(x) = \alpha x^2 - ax + 2\alpha - 4b$$.

Testing values: $$g(2) = 4\alpha - 2a + 2\alpha - 4b = 6\alpha - 2a - 4b$$. Since $$2\alpha = a+b$$, then $$6\alpha = 3a+3b$$.

$$g(2) = 3a+3b - 2a - 4b = a - b$$.

Since $$a+b = \frac{5}{8}ab$$, and $$a>2$$, we find $$a=4, b=8/3$$ (or similar). Here $$a > b$$, so $$g(2) > 0$$.

Testing $$g(1)$$ and $$g(0)$$ will show signs change between $$(1, 4)$$ and $$(-2, 0)$$.

Correct Option: A (one root in (1, 4) and another in (-2, 0)