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Step-by-Step Solution
Step 1: Identify the Arithmetic Mean
The arithmetic mean (A.M.) of two distinct real numbers $l$ and $n$ (both greater than 1) is given by
$m = \frac{l + n}{2}.
Step 2: Determine the Common Ratio of the Geometric Progression
We have three geometric means $G_1$, $G_2$, and $G_3$ between $l$ and $n$. This means that the sequence $l, G_1, G_2, G_3, n$ forms a geometric progression (G.P.) with common ratio $r$. Since there are four equal ratios from $l$ to $n$, we get
$r^4 = \frac{n}{l} \quad \Longrightarrow \quad r = \Bigl(\frac{n}{l}\Bigr)^{\tfrac{1}{4}}.
Step 3: Express the Three Geometric Means
The first term of this G.P. is $l$. The subsequent terms (the three geometric means) are:
$G_1 = l \cdot r = l \Bigl(\frac{n}{l}\Bigr)^{\tfrac{1}{4}} = l^{\tfrac{3}{4}} n^{\tfrac{1}{4}},$
$G_2 = l \cdot r^2 = l \Bigl(\frac{n}{l}\Bigr)^{\tfrac{1}{2}} = l^{\tfrac{1}{2}} n^{\tfrac{1}{2}},$
$G_3 = l \cdot r^3 = l \Bigl(\frac{n}{l}\Bigr)^{\tfrac{3}{4}} = l^{\tfrac{1}{4}} n^{\tfrac{3}{4}}.$
Step 4: Compute Each Required Power
We need to find $G_1^4 + 2\,G_2^4 + G_3^4.$ Calculate each term’s fourth power:
$G_1^4 = (l^{\tfrac{3}{4}} n^{\tfrac{1}{4}})^4 = l^3 n,$
$G_2^4 = (l^{\tfrac{1}{2}} n^{\tfrac{1}{2}})^4 = l^2 n^2,$
$G_3^4 = (l^{\tfrac{1}{4}} n^{\tfrac{3}{4}})^4 = l n^3.$
Step 5: Sum the Expressions
Add them together with the given coefficients:
$G_1^4 + 2\,G_2^4 + G_3^4 \;=\; l^3 n + 2 (\,l^2 n^2\,) + l n^3.
Factor out $l n$:
$l^3 n + 2 l^2 n^2 + l n^3 \;=\; l n \bigl[l^2 + 2 l n + n^2\bigr] = l n\, (l + n)^2.
Step 6: Relate $(l + n)$ to the Arithmetic Mean
Recall that $m = \frac{l + n}{2}$. Consequently, $(l + n) = 2m$ and thus $(l + n)^2 = 4 m^2.$
This gives:
$l n (l + n)^2 = l n \cdot 4 m^2 = 4\, l\, m^2\, n.
Step 7: State the Final Result
Hence, the value of $G_1^4 + 2G_2^4 + G_3^4$ is
$\boxed{4\,l\,m^2\,n}.
