Animated Solution for Mathematics - Sequence and Series: Let $a_1,a_2,\dots a_n$ be positive real numbers in geometric progression. For each $n$, let $A_n,G_n,H_n$ be respectively, the arithmetic mean, geometric mean, and harmonic mean of $a_1,a_2,\dots a_n$. Find an expression for the geometric mean of $G_1,G_2,\dots G_n$ in terms of $A_1,A_2,\dots A_n,H_1,H_2,\dots H_n$.

My dear student, welcome to a journey through one of the most elegant corners of algebra. Often, when we see terms like Arithmetic Mean ($A_n$), Geometric Mean ($G_n$), and Harmonic Mean ($H_n$) thrown together in a problem, our minds immediately jump to the AM-GM-HM inequality.

But today, we are going to do something different. We are not going to look for bounds; we are going to look for an identity. We are going to uncover the hidden harmony that exists when these means are generated by a Geometric Progression.

Let us begin by visualizing our sequence. We have $a_1,a_2,\dots ,a_n$, a sequence of positive real numbers in a Geometric Progression. The definition is simple: each term is the previous one multiplied by a common ratio $r$. So, the $k$-th term is $a_k=a_1{r}^{k-1}$.

Now, consider the first $k$ terms: $a_1,a_2,\dots ,a_k$. The Geometric Mean, $G_k$, is defined as the $k$-th root of their product:

Here is where the magic happens. In a G.P., there is a beautiful symmetry. If you multiply the first term and the last term, $a_1\cdot a_k$, you get the same result as multiplying the second term and the second-to-last term, $a_2\cdot a_{k-1}$. This product is constant!

Because we have $k$ terms, we can form $k/2$ such pairs. Thus, the product of all $k$ terms is simply $(a_1{a}_k{)}^{k/2}$. Substituting this back into our definition of $G_k$, we get:

This is our first pillar. The geometric mean of the first $k$ terms of a G.P. is just the square root of the product of the first and last terms. Keep this in your toolkit; it is the key to everything that follows.

Next, let us tackle the Harmonic Mean, $H_k$. By definition, $H_k$ is the reciprocal of the arithmetic mean of the reciprocals:

At first glance, this denominator looks intimidating. But remember, the reciprocals of a G.P. also form a G.P.! More importantly, look at the sum ${\sum }_{i=1}^k\frac{1}{a_i}$.

Using the same symmetry we discovered earlier, we can write $\frac{1}{a_i}=\frac{a_{k-i+1}}{a_1{a}_k}$. When we sum these up, the numerator becomes the sum of the original terms, ${\sum }_{i=1}^k{a}_i$, and the denominator is $a_1{a}_k$.

So, the sum of reciprocals is simply $\frac{\sum a_i}{a_1{a}_k}$. Substituting this into our expression for $H_k$:

Since the Arithmetic Mean $A_k$ is defined as $\frac{\sum a_i}{k}$, we can see that $\frac{k}{\sum a_i}=\frac{1}{A_k}$. Therefore:

Now, let us bring it all together. We have $G_k=\sqrt{a_1{a}_k}$, which implies $G_k^2=a_1{a}_k$. We also have $H_k=\frac{a_1{a}_k}{A_k}$, which implies $a_1{a}_k=A_k{H}_k$.

Equating these two, we arrive at the core identity for any G.P.:

This is the "Aha!" moment. For any $k$, the geometric mean is the geometric mean of the arithmetic and harmonic means.

Finally, the problem asks for the geometric mean of the sequence $G_1,G_2,\dots ,G_n$. Let us call this $G$. By definition:

Substitute our identity $G_k=(A_k{H}_k{)}^{1/2}$ into this expression:

Using the laws of exponents, we can combine the inner exponents of $1/2$ with the outer exponent of $1/n$. The result is elegant and clean:

And there you have it! We have expressed the geometric mean of the means in terms of the arithmetic and harmonic means. It is a beautiful result, isn't it? Mathematics is not just about solving for $x$; it is about finding the hidden connections between seemingly disparate concepts. Keep this curiosity alive, and you will conquer any problem JEE throws at you.