For x ≥ 0, the least value of K, for which 4^(1+x) + 4^(1-x), K/2, 16^x + 16^(-x) are three consecut

Q: Wait, why do we use the AM-GM inequality here? Can't we just use calculus to find the minimum?

You absolutely can use calculus! If you define $f(x) = 4^{1+x} + 4^{1-x} + 16^x + 16^{-x}$ and find $f'(x) = 0$, you will arrive at the same result. However, in JEE Advanced, time is your most precious resource. Recognizing the structure of $y + \frac{1}{y}$ and immediately applying AM-GM is a much faster, more elegant, and less error-prone method. It is a classic pattern that you should train your eyes to spot instantly.

Q: Do we need to check if the minimums for both parts occur at the same value of $x$?

That is a brilliant question! Yes, you must. If the first part reached its minimum at $x=1$ and the second part at $x=2$, you could not simply add the minimums together. In this problem, both $4^x + 4^{-x}$ and $16^x + 16^{-x}$ reach their minimum when $4^x = 1$ and $16^x = 1$, which both occur at $x=0$. Since they share the same minimizing $x$, adding the minimums is perfectly valid.

Q: What if the problem asked for the maximum value? Would AM-GM still work?

No, AM-GM only provides a lower bound (a minimum). For a maximum, you would need to look at the boundaries of the domain. Since $x \geq 0$, as $x$ increases, $4^x$ and $16^x$ grow without bound, meaning the expression would approach infinity. Therefore, a maximum does not exist in this domain.

Question

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 :

Options:

(A) 8

(B) 4

(C) 10

(D) 16

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a sequence of numbers. You are told they form an Arithmetic Progression (A.P.), a beautiful, rhythmic structure where the gap between consecutive terms is constant. Our mission is to find the least value of a mysterious constant $K$ hidden within these terms.

We are given three terms:
1. $a = 4^{1+x} + 4^{1-x}$
2. $b = \frac{K}{2}$
3. $c = 16^x + 16^{-x}$

For any three consecutive terms in an A.P., the middle term is the arithmetic mean of the extremes. This is our golden key: $2b = a + c$.

### The Master Equation

Substituting our terms into the A.P. condition, we get:

$$2\left(\frac{K}{2}\right) = \left(4^{1+x} + 4^{1-x}\right) + \left(16^x + 16^{-x}\right)$$

The $2$ on the left cancels out perfectly with the denominator. This leaves us with a clean, albeit intimidating, expression for $K$:

$$K = 4^{1+x} + 4^{1-x} + 16^x + 16^{-x}$$

### The Algebraic Transformation

Now, let us simplify. We know that $4^{1+x} = 4^1 \cdot 4^x$. The first part of our expression becomes $4 \cdot 4^x + 4 \cdot 4^{-x}$, which we can factor as $4(4^x + \frac{1}{4^x})$.

What about the second part, $16^x + 16^{-x}$? Notice that $16^x = (4^x)^2$. This reveals a hidden symmetry. Our equation now looks like this:

$$K = 4\left(4^x + \frac{1}{4^x}\right) + \left(16^x + \frac{1}{16^x}\right)$$

### The Power of AM-GM

Whenever you see a sum of a positive number and its reciprocal, your mathematical intuition should immediately trigger the **AM-GM inequality**: $\frac{y + \frac{1}{y}}{2} \geq \sqrt{y \cdot \frac{1}{y}}$, which simplifies to $y + \frac{1}{y} \geq 2$.

Let us apply this to our two chunks:
1. For the first part, let $y = 4^x$. Since $4^x > 0$, we have $4^x + \frac{1}{4^x} \geq 2$. Multiplying by $4$, we get $4(4^x + \frac{1}{4^x}) \geq 8$.
2. For the second part, let $y = 16^x$. Similarly, $16^x + \frac{1}{16^x} \geq 2$.

Adding these together, we find that $K \geq 8 + 2$, or $K \geq 10$.

### Final Calculation

We have found that $K$ must be at least $10$. We must verify if this minimum actually occurs. The AM-GM inequality holds equality only when the terms are equal.

For the first part, $4^x = \frac{1}{4^x} \Rightarrow 4^{2x} = 1 \Rightarrow x = 0$. For the second part, $16^x = \frac{1}{16^x} \Rightarrow 16^{2x} = 1 \Rightarrow x = 0$.

Since both parts reach their minimum at $x = 0$, the value $x = 0$ is perfectly valid. Thus, the **least value of $K$ is $10$**.

Animated Solution for Mathematics - Sequence and Series: For $x \geq 0$ , the least value of $K$ , for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 1 6^{x} + 1 6^{- x}$ are three consecutive terms of an A.P., is equal to :

Select Answer:

Visualized Solution

Identify the Terms of the A.P.

The Condition for A.P.

Substitute and Set Up Equation

Simplify Exponential Terms

Introduce AM-GM Inequality

Apply Inequality to First Part

Apply Inequality to Second Part

Calculate the Least Value of $K$

Final Answer

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Master Equation

The Algebraic Transformation

The Power of AM-GM

Final Calculation

Similar Questions

Let $f : R \to R$ be such that for all $x \in R$ , $(2^{1 + x} + 2^{1 - x})$ , $f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

Let $f : R \to R$ be such that for all $x \in R (2^{1 + x} + 2^{1 - x}), f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

The minimum value of the sum of real numbers $a^{- 5}, a^{- 4}, 3 a^{- 3}, 1, a^{8}$ and $a^{10}$ where $a > 0$ is .........

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then

The minimum value of $f (x) = a^{a^{x}} + a^{1 - a^{x}}$ , where $a, x \in R$ and $a > 0$ , is equal to:

Let $a, b, c$ be positive integers such that $b / a$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b + 2$ , then the value of $\frac{a ^{2} + a - 14}{a + 1}$ is \dots.

If $a_{1}, a_{2}, \dots, a_{n}$ are positive real numbers whose product is a fixed number $c$ , then the minimum value of $a_{1} + a_{2} + \dots + a_{n - 1} + 2 a_{n}$ is

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

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.