A) 9
B) 9/2
C) 27/4
D) 15/2
Correct Answer: C
Solution :
Infinite series \[9-3+1-\frac{1}{3}......\infty \] is a G.P. with \[a=9,r=\frac{-1}{3}\] \ \[{{S}_{\infty }}=\frac{a}{1-r}=\frac{9}{1+\left( \frac{1}{3} \right)}=\frac{9\times 3}{4}=\frac{27}{4}\].
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