Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric

means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to :

<p>Now, we have the following relations :</p> <p>Arithmetic progression :</p> <p>Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.</p> <p>To find the value of $d$, we can use the following equation :</p> <p>$b - a = 3d$</p> <p>From this equation, we can find the value of $d$ : </p> <p>$d = \frac{b - a}{3}$</p> <p>$A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3}$</p> <p>$A_2 = \frac{a + 2b}{3}$</p> <p>$A_1 + A_2 = a + b$</p> <p>Geometric progression :</p> <p>$a, G_1, G_2, G_3, b \text{ are in G.P. }$</p> <p>$r = \left(\frac{b}{a}\right)^{\frac{1}{4}}$</p> <p>$G_1 = \left(a^3b\right)^{\frac{1}{4}}$</p> <p>$G_2 = \left(a^2b^2\right)^{\frac{1}{4}}$</p> <p>$G_3 = \left(ab^3\right)^{\frac{1}{4}}$</p> <p>We have the expression :</p> <p>$$ G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}} $$</p> <p>Simplify the expression :</p> <p>$a^3b + a^2b^2 + ab^3 + ab(a^2b^2)$</p> <p>Factor out $ab$:</p> <p>$ab(a^2 + ab + b^2 + a^2b^2)$</p> <p>Combine the terms :</p> <p>$ab(a^2 + 2ab + b^2)$</p> <p>Rewrite the expression using the sum of squares :</p> <p>$ab(a + b)^2$</p> <p>Now, recall that $A_1 + A_2 = a + b$. Substitute this into the expression :</p> <p>$G_1 \cdot G_3 \cdot (A_1 + A_2)^2$</p>