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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 geometric progression

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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 geometric progression $(G.P.)$. Here,the first term $a = 4$ and the fifth term $ar^4 = \frac{1}{4}$. Substituting $a = 4$,we get $4r^4 = \frac{1}{4}$,which implies $r^4 = \frac{1}{16} = (\frac{1}{2})^4$. Thus,$r = \frac{1}{2}$ (considering the positive common ratio). The product of the three geometric means is $g_1 	imes g_2 	imes g_3 = (ar) 	imes (ar^2) 	imes (ar^3) = a^3 r^6$. Substituting the values,we get $4^3 	imes (\frac{1}{2})^6 = 64 	imes \frac{1}{64} = 1$. Alternatively,the product of $n$ geometric means between $a$ and $b$ is $(ab)^{n/2}$. Here,$a = 4, b = \frac{1}{4}, n = 3$. Product $= (4 	imes \frac{1}{4})^{3/2} = 1^{3/2} = 1$.

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

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