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(D) Let the three geometric means between $4$ and $\frac{1}{4}$ be $g_1, g_2, g_3$.Then $4, g_1, g_2, g_3, \frac{1}{4}$ form a $G.P.$Here,the first te

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(D) Let the three geometric means between $4$ and $\frac{1}{4}$ be $g_1, g_2, g_3$.Then $4, g_1, g_2, g_3, \frac{1}{4}$ form a $G.P.$Here,the first term $a = 4$ and the fifth term $ar^4 = \frac{1}{4}$.$4r^4 = \frac{1}{4} \Rightarrow r^4 = \frac{1}{16} = \left(\frac{1}{2}ight)^4$.Thus,$r = \frac{1}{2}$ (taking the positive common ratio).The product of the three geometric means is $g_1 	imes g_2 	imes g_3 = (ar) 	imes (ar^2) 	imes (ar^3) = a^3r^6$.Substituting the values: $a^3r^6 = (4)^3 	imes \left(\frac{1}{2}ight)^6 = 64 	imes \frac{1}{64} = 1$.Alternatively,the product of $n$ geometric means between $a$ and $b$ is $(ab)^{n/2}$. Here $a=4, b=1/4, n=3$,so $(4 	imes \frac{1}{4})^{3/2} = 1^{3/2} = 1$.

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

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