A) 4
B) 6
C) 12
D) - 6
Correct Answer: B
Solution :
Let \[{{G}_{1}},{{G}_{2}},{{G}_{3}},{{G}_{4}},{{G}_{5}}\] be the G.M.?s are inserted between 486 and 2/3. So total terms are 7. \[{{T}_{n}}=a{{r}^{n-1}}\] Þ 2/3 = 486\[{{(r)}^{6}}\]\[\Rightarrow r=1/3\] Hence 4th G.M. will be, \[{{T}_{5}}=a{{r}^{4}}=486\,{{(1/3)}^{4}}=6\].
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