Let A_1 and A_2 be two arithmetic means and G_1, G_2, G_3 be three geometric means of two …

Question

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 :

Accepted Answer

Now, we have the following relations :

Arithmetic progression :

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 $d$.

To find the value of $d$, we can use the following equation :

$b - a = 3d$

From this equation, we can find the value of $d$ :

$$
d = \frac{b - a}{3}
$$

$$
A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3}
$$

$$
A_2 = \frac{a + 2b}{3}
$$

$A_1 + A_2 = a + b$

Geometric progression :

$$
a, G_1, G_2, G_3, b \text{ are in G.P. }
$$

$$
r = \left(\frac{b}{a}\right)^{\frac{1}{4}}
$$

$$
G_1 = \left(a^3b\right)^{\frac{1}{4}}
$$

$$
G_2 = \left(a^2b^2\right)^{\frac{1}{4}}
$$

$$
G_3 = \left(ab^3\right)^{\frac{1}{4}}
$$

We have the expression :

$$
G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}}
$$

Simplify the expression :

$a^3b + a^2b^2 + ab^3 + ab(a^2b^2)$

Factor out $ab$:

$ab(a^2 + ab + b^2 + a^2b^2)$

Combine the terms :

$ab(a^2 + 2ab + b^2)$

Rewrite the expression using the sum of squares :

$ab(a + b)^2$

Now, recall that $A_1 + A_2 = a + b$. Substitute this into the expression :

$$
G_1 \cdot G_3 \cdot (A_1 + A_2)^2
$$

Solutions · 1