For $$x \geq 0$$, the least value of $$K$$, for which $$4^{1+x} + 4^{1-x}, \frac{K}{2}, 16^x + 16^{-x}$$ are three consecutive terms of an A.P., is equal to :

We need $$4^{1+x} + 4^{1-x}$$, $$\frac{K}{2}$$ and $$16^x + 16^{-x}$$ to be three consecutive terms of an A.P., which means the middle term equals the average of the other two:

Let $$t = 4^x$$ where $$t \geq 1$$ since $$x \geq 0$$. Then $$4^{1+x} = 4\cdot4^x = 4t$$, $$4^{1-x} = \frac{4}{t}$$, $$16^x = (4^x)^2 = t^2$$ and $$16^{-x} = \frac{1}{t^2}$$, so

To find the minimum of $$K$$ let $$u = t + \frac{1}{t}$$. Since $$t \geq 1$$, by AM-GM we have $$u \geq 2$$. Noting that $$t^2 + \frac{1}{t^2} = u^2 - 2$$ and $$4t + \frac{4}{t} = 4u$$, it follows that

As a quadratic in $$u$$, this opens upwards with vertex at $$u = -2$$, and hence is increasing for $$u \geq 2$$. Therefore the minimum occurs at $$u = 2$$, corresponding to $$t = 1$$ and $$x = 0$$, giving

The correct answer is Option (3): 10.