Question
For the following G.P.s, find $S_n$. 3, 6, 12, 24, ……..
Here, $a=3, r=\frac{6}{3}=2>1$
$\begin{aligned} & S_n=\frac{a\left(r^n-1\right)}{r-1}, \text { for } r>1 \\ \therefore & S_n=\frac{3\left(2^n-1\right)}{2-1} \\ \therefore & S_n=3\left(2^n-1\right)\end{aligned}$
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A’ ∩ B