Let A_1 and A_2 be two arithmetic means and G | vedclass

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(B) Let the two numbers be $a$ and $b$.$A_1, A_2$ are arithmetic means between $a$ and $b$,so $a, A_1, A_2, b$ are in $A$.$P$.Common difference $d = \

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$(A_1 + A_2)^2 G_1 G_3$

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(B) Let the two numbers be $a$ and $b$.$A_1, A_2$ are arithmetic means between $a$ and $b$,so $a, A_1, A_2, b$ are in $A$.$P$.Common difference $d = \frac{b-a}{3}$.$A_1 = a + \frac{b-a}{3} = \frac{2a+b}{3}$ and $A_2 = a + \frac{2(b-a)}{3} = \frac{a+2b}{3}$.Thus,$A_1 + A_2 = a + b$.$G_1, G_2, G_3$ are geometric means between $a$ and $b$,so $a, G_1, G_2, G_3, b$ are in $G$.$P$.Common ratio $r = (b/a)^{1/4}$.$G_1 = ar = a(b/a)^{1/4} = a^{3/4}b^{1/4}$,$G_2 = ar^2 = a^{1/2}b^{1/2}$,$G_3 = ar^3 = a^{1/4}b^{3/4}$.$G_1^4 = a^3b$,$G_2^4 = a^2b^2$,$G_3^4 = ab^3$.$G_1^2 G_3^2 = (a^{3/2}b^{1/2})(a^{1/2}b^{3/2}) = a^2b^2$.Sum $= G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + a^2b^2 = a^3b + 2a^2b^2 + ab^3 = ab(a^2 + 2ab + b^2) = ab(a+b)^2$.Since $G_1 G_3 = (a^{3/4}b^{1/4})(a^{1/4}b^{3/4}) = ab$,the expression becomes $(A_1 + A_2)^2 G_1 G_3$.

Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2, G_3$ be three geometric means between two distinct positive numbers $a$ and $b$. Then $G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$ is equal to

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