A) 5
B) 6
C) 7
D) 8
Correct Answer: B
Solution :
\[{{a}_{1}}=3,\ {{a}_{n}}=96\]\[\Rightarrow \]\[{{a}_{1}}{{r}^{n-1}}=96\]\[\Rightarrow \]\[{{r}^{n-1}}=32\] ?..(i) \[{{S}_{n}}=\frac{{{a}_{1}}({{r}^{n}}-1)}{r-1}=189\]\[\Rightarrow \]\[\frac{3(32r-1)}{r-1}=189\] Hence \[r=2\] and\[n=6\].
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