Key Concepts and Formulas

• Arithmetic Progression (A.P.): Three numbers $a, b, c$ are in A.P. if $b - a = c - b$, which simplifies to $2b = a + c$.
• Exponent Rules: $a^{m+n} = a^m \cdot a^n$, $a^{m-n} = a^m \cdot a^{-n}$, $(a^m)^n = a^{mn}$.
• AM-GM Inequality: For non-negative numbers $p_1, p_2, \dots, p_n$, $\frac{p_1 + p_2 + \dots + p_n}{n} \ge \sqrt[n]{p_1 p_2 \dots p_n}$. For two non-negative numbers $a, b$, $\frac{a+b}{2} \ge \sqrt{ab}$. Equality holds if and only if $a=b$.
• Minimum of $P + P^{-1}$: For $P > 0$, $P + P^{-1} \ge 2$, with equality when $P=1$.

Step-by-Step Solution

Step 1: Understand the Problem and Apply the A.P. Condition The problem states that three expressions are consecutive terms of an Arithmetic Progression (A.P.). The given terms are $a = 4^{1+x} + 4^{1-x}$, $b = \frac{\mathrm{K}}{2}$, and $c = 16^x + 16^{-x}$. For these to be in A.P., the middle term must be the average of the other two, i.e., $2b = a+c$. Substituting the given terms into the A.P. condition: $2 \left( \frac{\mathrm{K}}{2} \right) = (4^{1+x} + 4^{1-x}) + (16^x + 16^{-x})$ This simplifies to: $\mathrm{K} = (4^{1+x} + 4^{1-x}) + (16^x + 16^{-x})$ Our goal is to find the least value of $\mathrm{K}$ for $x \ge 0$.

Step 2: Simplify the Exponential Expressions To analyze the expression for $\mathrm{K}$, we simplify the terms involving exponents by expressing them with a common base, preferably $2$.

For the first term, $4^{1+x} + 4^{1-x}$: Using $4 = 2^2$, we have: $4^{1+x} = 4^1 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x}$ $4^{1-x} = 4^1 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x}$ So, the first term becomes: $4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})$

For the third term, $16^x + 16^{-x}$: Using $16 = 2^4$, we have: $16^x = (2^4)^x = 2^{4x}$ $16^{-x} = (2^4)^{-x} = 2^{-4x}$ So, the third term becomes: $16^x + 16^{-x} = 2^{4x} + 2^{-4x}$

Step 3: Substitute Simplified Terms and Form the Expression for K Substitute the simplified expressions back into the equation for $\mathrm{K}$: $\mathrm{K} = 4(2^{2x} + 2^{-2x}) + (2^{4x} + 2^{-4x})$ We need to find the minimum value of this expression for $x \ge 0$.

Step 4: Apply the AM-GM Inequality to Find the Minimum Value We can use the AM-GM inequality in the form $P + P^{-1} \ge 2$ for $P > 0$.

Consider the term $2^{2x} + 2^{-2x}$. Let $P = 2^{2x}$. Since $x \ge 0$, $2^{2x}$ is always positive. By AM-GM, $2^{2x} + 2^{-2x} \ge 2$. The equality holds when $2^{2x} = 1$, which means $2x = 0$, so $x=0$.

Consider the term $2^{4x} + 2^{-4x}$. Let $P = 2^{4x}$. Since $x \ge 0$, $2^{4x}$ is always positive. By AM-GM, $2^{4x} + 2^{-4x} \ge 2$. The equality holds when $2^{4x} = 1$, which means $4x = 0$, so $x=0$.

To minimize the entire expression for $\mathrm{K}$, we need to find a value of $x$ where both $2^{2x} + 2^{-2x}$ and $2^{4x} + 2^{-4x}$ achieve their minimum values simultaneously. As shown, both minimums occur at $x=0$. Since the domain is $x \ge 0$, $x=0$ is a valid value.

Step 5: Calculate the Least Value of K The least value of $\mathrm{K}$ occurs when $x=0$. Substitute $x=0$ into the expression for $\mathrm{K}$: $\mathrm{K}_{\text{least}} = 4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + (2^{4 \cdot 0} + 2^{-4 \cdot 0})$ $\mathrm{K}_{\text{least}} = 4(2^0 + 2^0) + (2^0 + 2^0)$ $\mathrm{K}_{\text{least}} = 4(1 + 1) + (1 + 1)$ $\mathrm{K}_{\text{least}} = 4(2) + 2$ $\mathrm{K}_{\text{least}} = 8 + 2$ $\mathrm{K}_{\text{least}} = 10$

Common Mistakes & Tips

• Not verifying the minimum: Simply substituting $x=0$ without using AM-GM or calculus to prove it yields the least value might be incomplete. The AM-GM condition for equality ensures that $x=0$ is indeed the point of minimum.
• Incorrectly applying AM-GM: Ensure that the terms to which AM-GM is applied are non-negative. In this case, $2^{2x}$ and $2^{4x}$ are always positive for real $x$.
• Algebraic errors with exponents: Double-check the simplification of exponential terms, especially when dealing with negative exponents or powers of powers.

Summary The problem requires us to find the least value of $\mathrm{K}$ such that three given expressions form an Arithmetic Progression. By applying the A.P. condition, we obtained an expression for $\mathrm{K}$ in terms of $x$. We then simplified the exponential terms and used the AM-GM inequality to determine the minimum value of the expression. The minimum value of $P + P^{-1}$ is $2$, occurring when $P=1$. Both terms in the expression for $\mathrm{K}$ achieved their minimums simultaneously at $x=0$. Substituting $x=0$ into the expression for $\mathrm{K}$ yielded the least value.

The final answer is $\boxed{10}$.