If a is the arithmetic mean of b and c,and G_ | vedclass

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(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,so $b+c = 2a$.Since $G_1, G_2$ are two geometric means between $b

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$2 G_1 G_2 a$

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(B) Given that $a$ is the arithmetic mean of $b$ and $c$,we have $a = \frac{b+c}{2}$,so $b+c = 2a$.Since $G_1, G_2$ are two geometric means between $b$ and $c$,the sequence $b, G_1, G_2, c$ is in Geometric Progression $(GP)$.Let the common ratio be $r$. Then $G_1 = br$,$G_2 = br^2$,and $c = br^3$.Thus,$r = (c/b)^{1/3}$.$G_1 = b(c/b)^{1/3} = b^{2/3} c^{1/3}$ and $G_2 = b(c/b)^{2/3} = b^{1/3} c^{2/3}$.Now,$G_1^3 + G_2^3 = (b^{2/3} c^{1/3})^3 + (b^{1/3} c^{2/3})^3 = b^2 c + b c^2 = bc(b+c)$.Substituting $b+c = 2a$,we get $G_1^3 + G_2^3 = bc(2a)$.Since $G_1 G_2 = (b^{2/3} c^{1/3})(b^{1/3} c^{2/3}) = bc$,we have $G_1^3 + G_2^3 = 2 a G_1 G_2$.

If $a$ is the arithmetic mean of $b$ and $c$,and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

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