Key Concepts and Formulas

• Arithmetic Mean (AM): For two numbers $a$ and $b$, their AM is $m = \frac{a+b}{2}$.
• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $k$-th term is given by $a_k = a_1 \cdot r^{k-1}$.
• Geometric Means (GMs): If $k$ geometric means ($G_1, G_2, \dots, G_k$) are inserted between two numbers $l$ and $n$, then $l, G_1, G_2, \dots, G_k, n$ form a G.P. The common ratio $r$ is given by $r = \left(\frac{n}{l}\right)^{\frac{1}{k+1}}$.

Step-by-Step Solution

Step 1: Understand the Given Information and Formulate Equations

We are given two distinct real numbers $l$ and $n$, both greater than 1 ($l, n > 1$).

• The arithmetic mean of $l$ and $n$ is $m$. By definition of AM: $m = \frac{l+n}{2}$ This implies: $l+n = 2m$

  • Why this step: This equation establishes a direct relationship between $l, n,$ and $m$. This will be essential for expressing our final answer in the required format.

• $G_1, G_2, G_3$ are three geometric means inserted between $l$ and $n$. This means the sequence $l, G_1, G_2, G_3, n$ forms a Geometric Progression (GP).

  • Why this step: Recognizing this as a GP allows us to use the properties of GPs to find the common ratio and subsequently the values of the geometric means.
  • In this GP, $l$ is the first term, and $n$ is the $(3+2)^{th} = 5^{th}$ term.

Step 2: Determine the Common Ratio ($r$) of the Geometric Progression

For a GP where $k$ geometric means are inserted between $l$ and $n$, the common ratio $r$ is given by: $r = \left(\frac{n}{l}\right)^{\frac{1}{k+1}}$ In this problem, $k=3$. Substituting this value: $r = \left(\frac{n}{l}\right)^{\frac{1}{3+1}} = \left(\frac{n}{l}\right)^{\frac{1}{4}}$

• Why this step: The common ratio is fundamental to defining all terms in a GP. Finding $r$ in terms of $l$ and $n$ enables us to express the geometric means ($G_1, G_2, G_3$) explicitly.

Step 3: Express the Geometric Means ($G_1, G_2, G_3$) in terms of $l$ and $n$

The terms of the GP are $l, l \cdot r, l \cdot r^2, l \cdot r^3, l \cdot r^4, \dots$. Therefore, the geometric means are:

• $G_1 = l \cdot r = l \cdot \left(\frac{n}{l}\right)^{\frac{1}{4}} = l^1 \cdot n^{\frac{1}{4}} \cdot l^{-\frac{1}{4}} = l^{\frac{3}{4}} n^{\frac{1}{4}}$
• $G_2 = l \cdot r^2 = l \cdot \left(\left(\frac{n}{l}\right)^{\frac{1}{4}}\right)^2 = l \cdot \left(\frac{n}{l}\right)^{\frac{2}{4}} = l \cdot \left(\frac{n}{l}\right)^{\frac{1}{2}} = l^1 \cdot n^{\frac{1}{2}} \cdot l^{-\frac{1}{2}} = l^{\frac{1}{2}} n^{\frac{1}{2}}$
• $G_3 = l \cdot r^3 = l \cdot \left(\left(\frac{n}{l}\right)^{\frac{1}{4}}\right)^3 = l \cdot \left(\frac{n}{l}\right)^{\frac{3}{4}} = l^1 \cdot n^{\frac{3}{4}} \cdot l^{-\frac{3}{4}} = l^{\frac{1}{4}} n^{\frac{3}{4}}$
• Why this step: These explicit expressions for $G_1, G_2,$ and $G_3$ are necessary for substitution into the expression we need to evaluate ($G_1^4 + 2G_2^4 + G_3^4$).

Step 4: Calculate the Fourth Powers of the Geometric Means

We need to find $G_1^4, G_2^4,$ and $G_3^4$:

• $G_1^4 = \left(l^{\frac{3}{4}} n^{\frac{1}{4}}\right)^4 = l^{(\frac{3}{4} \cdot 4)} n^{(\frac{1}{4} \cdot 4)} = l^3 n^1 = l^3 n$
• $G_2^4 = \left(l^{\frac{1}{2}} n^{\frac{1}{2}}\right)^4 = l^{(\frac{1}{2} \cdot 4)} n^{(\frac{1}{2} \cdot 4)} = l^2 n^2$
• $G_3^4 = \left(l^{\frac{1}{4}} n^{\frac{3}{4}}\right)^4 = l^{(\frac{1}{4} \cdot 4)} n^{(\frac{3}{4} \cdot 4)} = l^1 n^3 = l n^3$
• Why this step: Raising these terms to the fourth power simplifies the fractional exponents, making the subsequent algebraic manipulation much easier.

Step 5: Substitute and Simplify the Expression $G_1^4 + 2G_2^4 + G_3^4$

Now, substitute the calculated fourth powers into the given expression: $G_1^4 + 2G_2^4 + G_3^4 = (l^3 n) + 2(l^2 n^2) + (l n^3)$ $= l^3 n + 2l^2 n^2 + l n^3$ We can factor out the common term $ln$: $= ln(l^2 + 2ln + n^2)$ The expression inside the parenthesis is a perfect square trinomial, $(l+n)^2$: $= ln(l+n)^2$

• Why this step: This algebraic simplification is crucial. Factoring and recognizing the perfect square allows us to connect the expression to the sum $(l+n)$, which we already know is related to $m$.

Step 6: Express the Result in Terms of $l, m,$ and $n$

From Step 1, we have the relationship $l+n = 2m$. Substitute this into the simplified expression: $ln(l+n)^2 = ln(2m)^2$ $= ln(4m^2)$ Rearranging the terms to match the option format: $= 4lm^2n$

• Why this step: This final substitution uses the relationship with the arithmetic mean $m$ to express the result in the form presented in the answer options.

Comparing our result $4lm^2n$ with the given options, we find it matches option (A).

Common Mistakes and Tips

• Common Ratio Calculation: Ensure the common ratio for $k$ geometric means between $l$ and $n$ is correctly calculated as $r = (n/l)^{1/(k+1)}$. A common error is using $1/k$.
• Exponent Arithmetic: Be careful with exponent rules, especially when dealing with powers of powers and products of powers. $(x^a)^b = x^{ab}$ and $x^a \cdot x^b = x^{a+b}$.
• Algebraic Identities: Recognizing algebraic identities, such as $(l+n)^2 = l^2 + 2ln + n^2$, can significantly simplify the problem.
• Symmetry in Geometric Means: For $k$ GMs between $l$ and $n$, the product of the $i$-th GM from the beginning and the $i$-th GM from the end is equal to the product of the extreme terms, i.e., $G_i \cdot G_{k-i+1} = ln$. In this case, $G_1 \cdot G_3 = ln$. Also, $G_2 = \sqrt{ln}$.

Summary

We began by defining the arithmetic mean $m$ and establishing $l+n=2m$. Then, we identified that inserting three geometric means ($G_1, G_2, G_3$) between $l$ and $n$ creates a geometric progression. We calculated the common ratio $r = (n/l)^{1/4}$ and expressed each geometric mean in terms of $l$ and $n$. By calculating their fourth powers ($G_1^4=l^3n$, $G_2^4=l^2n^2$, $G_3^4=ln^3$) and substituting them into the expression $G_1^4 + 2G_2^4 + G_3^4$, we simplified it to $ln(l+n)^2$. Finally, using the relationship $l+n=2m$, we arrived at the answer $4lm^2n$.

The final answer is \boxed{4,lm{n^2}}.