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

<p>To determine the least value of $\mathrm{K}$ for which the terms $4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.</p> <p>For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:</p> <p>$\frac{4^{1+x} + 4^{1-x} + 16^x + 16^{-x}}{2} = \frac{K}{2}$</p> <p>First, simplify each term individually:</p> <p>1. Consider $4^{1+x} + 4^{1-x}$:</p> <p>$4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x} = 4 \cdot 2^{2x}$</p> <p>and</p> <p>$$4^{1-x} = 4 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} = 4 \cdot 2^{-2x}$$</p> <p>Thus,</p> <p>$4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})$</p> <p>2. Consider $16^x + 16^{-x}$:</p> <p>$16^x = (2^4)^x = 2^{4x}$</p> <p>and</p> <p>$16^{-x} = (2^4)^{-x} = 2^{-4x}$</p> <p>Thus,</p> <p>$16^x + 16^{-x} = 2^{4x} + 2^{-4x}$</p> <p>3. Combine the terms and set up the equation:</p> <p>$\frac{4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x}}{2} = \frac{K}{2}$</p> <p>Multiply both sides by 2:</p> <p>$4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x} = K$</p> <p>To find the least value of $\mathrm{K}$, let's assume $x = 0$ (since $x$ can range over non-negative values):</p> <p>For $x = 0$:</p> <p>$4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + 2^{4 \cdot 0} + 2^{-4 \cdot 0}$</p> <p>This simplifies to:</p> <p>$4(2^0 + 2^0) + 2^0 + 2^0$</p> <p>$= 4(1 + 1) + 1 + 1$</p> <p>$= 4 \cdot 2 + 1 + 1$</p> <p>$= 8 + 1 + 1$</p> <p>$= 10$</p> <p>Therefore, the least value of $\mathrm{K}$ that ensures the values form an arithmetic progression is $ 10 $. Hence, the correct option is:</p> <p>Option A: 10</p>