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 :

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Answer: A. \left(A_{1}+A_{2}ight)^{2} G_{1} G_{3} 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 	ext{ are in G.P. } r = \left(\frac{b}{a}ight)^{\frac{1}{4}} G_1 = \left(a^3bight)^{\frac{1}{4}} G_2 = \left(a^2b^2ight)^{\frac{1}{4}} G_3 = \left(ab^3ight)^{\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ight)^{\frac{1}{2}}\cdot\left(ab^3ight)^{\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

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 :

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