A) \[2\]
B) \[\frac{5}{2}\]
C) \[\frac{7}{2}\]
D) \[\frac{3}{2}\]
E) \[\frac{9}{2}\]
Correct Answer: D
Solution :
According to question \[1=2(r+{{r}^{2}}+{{r}^{3}}+....)\] \[\frac{1}{2}=\frac{r}{1-r}\] \[\Rightarrow \] \[r=\frac{1}{3}\] \[\therefore \]The series is \[1,\frac{1}{3},\frac{1}{9},\frac{1}{27},......\] Required sum\[=1+\frac{1}{3}+\frac{1}{9}+.....\] \[=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\]
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