Animated Solution for Mathematics - Sequence and Series: If $m$ is the A.M. of two distinct real numbers $l$ and $n$ ($l,n>1$) and $G_1,G_2$ and $G_3$ are three geometric means between $l$ and $n$, then $G_1^4+2G_2^4+G_3^4$ equals:

A Journey Through Sequences

Welcome, future engineer. Today, we are not just solving an algebra problem; we are exploring the elegant architecture of sequences. When we talk about Arithmetic Means (A.M.) and Geometric Means (G.M.), we are talking about the two fundamental ways to bridge the gap between two numbers.

Let us embark on this journey to uncover the hidden symmetry in the expression $G_1^4+2G_2^4+G_3^4$.

The Anchor of Arithmetic

Imagine you are standing on a number line. You have two points, $l$ and $n$. We are told that $m$ is the Arithmetic Mean of these two.

This is our anchor. By definition:

This is the midpoint, the balance point. But in the heat of a JEE exam, don't just look at this as a definition. Look at it as a tool.

If $m=\frac{l+n}{2}$, then $l+n=2m$. This simple rearrangement is our secret weapon. Keep it in your pocket; we will need it for the grand finale.

The Geometric Bridge

Now, let us insert three geometric means, $G_1,G_2,$ and $G_3$, between $l$ and $n$. This creates a sequence: $l,G_1,G_2,G_3,n$. This is a Geometric Progression (G.P.).

In a G.P., every step is a multiplication by a common ratio, $r$. Think about the journey from $l$ to $n$:

- To get to $G_1$, we multiply by $r$. - To get to $G_2$, we multiply by $r^2$. - To get to $G_3$, we multiply by $r^3$. - To get to $n$, we multiply by $r^4$.

So, $n=l\cdot r^4$. This is the key to unlocking the common ratio. Solving for $r$, we get:

This might look intimidating with those fractional exponents, but stay calm. The beauty of mathematics is that complexity often collapses into simplicity.

The Algebraic Dance

We need to calculate $G_1^4,G_2^4,$ and $G_3^4$. Let us take them one by one.

For $G_1=l\cdot r$, raising it to the fourth power gives $G_1^4=l^4\cdot r^4$. Since $r^4=\frac{n}{l}$, we have:

For $G_2=l\cdot r^2$, raising it to the fourth power gives $G_2^4=l^4\cdot r^8$. Since $r^4=\frac{n}{l}$, then $r^8=(r^4{)}^2=\frac{n^2}{l^2}$. Thus:

For $G_3=l\cdot r^3$, raising it to the fourth power gives $G_3^4=l^4\cdot r^{12}$. Since $r^{12}=(r^4{)}^3=\frac{n^3}{l^3}$. Thus:

Look at the pattern! We have $l^{3}n$, $l^2{n}^2$, and $l{n}^3$. The exponents are shifting with such grace.

The Grand Finale

Now, let us assemble the target expression: $G_1^4+2G_2^4+G_3^4$. Substituting our findings, we get:

Do you see it? Every term shares a common factor of $ln$. Let us pull it out:

That expression inside the parentheses is a classic identity: $(l+n{)}^2$. So, we have $ln(l+n{)}^2$.

Finally, remember our secret weapon from Phase 1? $l+n=2m$. Let us substitute that in:

And there it is. The complexity has vanished, leaving behind a clean, elegant result. This is the essence of JEE Advanced mathematics—not brute force, but the art of seeing the structure beneath the surface. You have mastered the sequence.