A) 9
B) 9/2
C) 27/4
D) 15/2
Correct Answer: C
Solution :
Given series is \[9-3+1-\frac{1}{3}+\,.......\,\,\infty \] This is an infinite G.P. series, here \[a=9,\,\,\,r=-\frac{1}{3}\] \[{{S}_{\infty }}=\frac{a}{1-r}=\frac{9}{1-\left( -\frac{1}{3} \right)}\] \[=\frac{9}{\frac{4}{3}}=\frac{27}{4}\]You need to login to perform this action.
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