A) 4
B) 2
C) \[-1\]
D) 1
Correct Answer: D
Solution :
We have \[4,\ {{g}_{1}},\ {{g}_{2}},\ {{g}_{3}},\ \frac{1}{4}\] is a G.P. Here\[a=4\], \[{{g}_{1}}=ar=4\times r,\ \ \ {{g}_{2}}=a{{r}^{2}},\ \ \ {{g}_{3}}=a{{r}^{3}}\] and \[{{g}_{4}}=a{{r}^{4}}=4\times {{r}^{4}}=\frac{1}{4}\] \[\Rightarrow {{r}^{4}}=\frac{1}{16}={{\left( \frac{1}{2} \right)}^{4}}\Rightarrow r=\frac{1}{2}\] Now product of three G.M. \[{{g}_{1}}.\ {{g}_{2}}.\ {{g}_{3}}=ar.\ a{{r}^{2}}.\ a{{r}^{3}}\] \[={{a}^{3}}{{r}^{6}}={{4}^{3}}\times {{\left( \frac{1}{2} \right)}^{6}}=\frac{{{4}^{3}}}{{{4}^{3}}}=1\]. Note: The product of ?\[n\]? geometric means between ?\[a\]? and \[\frac{1}{a}\] is always equal to 1.
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