Let $$A_1$$ and $$A_2$$ be two arithmetic means and $$G_1$$, $$G_2$$ and $$G_3$$ 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

Let the two distinct positive numbers be $$a$$ and $$b$$.

When two arithmetic means $$A_1$$ and $$A_2$$ are inserted between $$a$$ and $$b$$, the sequence $$a, A_1, A_2, b$$ is in AP. The common difference is $$d = \frac{b-a}{3}$$, so:

$$A_1 = a + d$$ and $$A_2 = a + 2d$$

$$A_1 + A_2 = 2a + 3d = 2a + 3 \cdot \frac{b-a}{3} = 2a + b - a = a + b$$

When three geometric means $$G_1, G_2, G_3$$ are inserted between $$a$$ and $$b$$, the sequence $$a, G_1, G_2, G_3, b$$ is in GP with common ratio $$r = \left(\frac{b}{a}\right)^{1/4}$$. Therefore:

$$G_1 = ar = a\left(\frac{b}{a}\right)^{1/4} = a^{3/4}b^{1/4}$$ $$G_2 = ar^2 = a\left(\frac{b}{a}\right)^{1/2} = a^{1/2}b^{1/2} = \sqrt{ab}$$ $$G_3 = ar^3 = a\left(\frac{b}{a}\right)^{3/4} = a^{1/4}b^{3/4}$$ $$G_1 G_3 = a^{3/4}b^{1/4} \times a^{1/4}b^{3/4} = ab$$ $$G_1^2 = a^{3/2}b^{1/2}, \quad G_2^2 = ab, \quad G_3^2 = a^{1/2}b^{3/2}$$ $$G_1^4 = a^3b, \quad G_2^4 = a^2b^2, \quad G_3^4 = ab^3$$ $$G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$$ $$= a^3b + a^2b^2 + ab^3 + G_1^2 G_3^2$$

Now, $$G_1^2 G_3^2 = (G_1 G_3)^2 = (ab)^2 = a^2b^2$$. So:

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

From Steps 1 and 3: $$A_1 + A_2 = a + b$$ and $$G_1 G_3 = ab$$. Therefore:

$$ab(a+b)^2 = G_1 G_3 \cdot (A_1 + A_2)^2 = (A_1 + A_2)^2 G_1 G_3$$

The correct answer is Option 1: $$(A_1 + A_2)^2 G_1 G_3$$.