Animated Solution for Mathematics - Sequence and Series: Comprehension Passage

Imagine you are standing on a number line, holding two distinct positive numbers, $a$ and $b$. Let us assume $a<b$. You are about to embark on a recursive journey, creating a sequence of means that will dance toward each other.

We begin with the fundamental trio: the Arithmetic Mean $A_1=\frac{a+b}{2}$, the Geometric Mean $G_1=\sqrt{ab}$, and the Harmonic Mean $H_1=\frac{2ab}{a+b}$.

For any two distinct positive numbers, we have the ironclad inequality $A_1>G_1>H_1$. This is our starting point, our anchor. The arithmetic mean sits to the right, the harmonic mean to the left, and the geometric mean rests in the middle.

Now, we enter the loop. For $n\ge 2$, we define the next generation of means from the previous generation:

It is a feedback loop, where the outputs of one step become the inputs of the next.

Let us calculate $G_2$. By definition, $G_2=\sqrt{A_1{H}_1}$.

Here is where the magic happens. Recall the beautiful identity $A_1{H}_1=G_1^2$. When we substitute this into our equation, we get:

The geometric mean is invariant; it does not change. Because the rule is recursive, this holds for all $n$. Thus, $G_1=G_2=G_3=\dots$. The geometric mean is the fixed, unmoving center of our system.

Now, consider the arithmetic mean $A_n$. To see how it behaves, we examine the difference:

Since we know $A_{n-1}>H_{n-1}$, this difference is strictly negative. This means $A_n<A_{n-1}$.

With every step, the arithmetic mean is being pulled down, closer to the geometric mean. It forms a strictly decreasing sequence: $A_1>A_2>A_3>\dots$.

Finally, we look at the harmonic mean $H_n$. We use our invariant property $A_n{H}_n=G_1^2$, which gives us:

Since $A_n$ is decreasing, its reciprocal $\frac{1}{A_n}$ is increasing. Therefore, $H_n$ must be strictly increasing: $H_1<H_2<H_3<\dots$. The harmonic mean is being pushed up, climbing toward the geometric mean.

We have witnessed a beautiful phenomenon. The arithmetic mean descends, the harmonic mean ascends, and both are relentlessly squeezed toward the constant geometric mean $G_1$.

They are marching toward a common destination, a perfect meeting point defined by the initial numbers. You have just solved a problem of convergence, proving that even in a complex recursive system, there is an underlying, elegant order.