A) \[\frac{1}{3}\]
B) \[\frac{4}{9}\]
C) \[\frac{2}{3}\]
D) \[\frac{2}{9}\]
Correct Answer: C
Solution :
Let a be the first term and r be the common ratio of the G.P. \[\therefore \]\[\frac{a}{1-r}=3\] \[\Rightarrow \]\[a=3(1-r)\] ?(i) Now, sum of the cubes of its terms is \[\frac{{{a}^{3}}}{1-{{r}^{3}}}=\frac{27}{19}\] \[\Rightarrow \]\[\frac{27{{(1-r)}^{3}}}{1-{{r}^{3}}}=\frac{27}{19}\] [From (i)] \[\Rightarrow \]\[\frac{(1-r)(1+{{r}^{2}}-2r)}{(1-r)(1+r+{{r}^{2}})}=\frac{1}{19}\] \[\Rightarrow \]\[19+19{{r}^{2}}-38r=1+r+{{r}^{2}}\] \[\Rightarrow \]\[6{{r}^{2}}-13r+6=0\]\[\Rightarrow \]\[r=\frac{3}{2}\] or \[\frac{2}{3}\] Since,\[|r|<1,\]so \[r=\frac{2}{3}.\]
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