A) 8
B) 4
C) 16
D) 12
Correct Answer: C
Solution :
\[2,\,{{g}_{1}},\ {{g}_{2}},\ {{g}_{3}},\ 32\]where \[a=2,\ ar={{g}_{1}},\ a{{r}^{2}}={{g}_{2}},\ a{{r}^{3}}={{g}_{3}}\] and\[a{{r}^{4}}=32\] Now\[2\times {{r}^{4}}=32\Rightarrow {{r}^{4}}=16={{(2)}^{4}}\Rightarrow r=2\]. Then third geometric mean \[=a{{r}^{3}}=2\times {{2}^{3}}=16\]. Aliter : By formula, \[{{G}_{3}}=2.{{\left( \frac{32}{2} \right)}^{3/4}}=2\ .\ 8=16\].
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