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The sum of a geometric progression with infinitely many terms is given by $s=\dfrac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the common ratio. We have

$r=\dfrac{a_2}{a_1}=\dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2}$

Thus,

$s=\dfrac{1}{1-\dfrac{1}{2}}=$ $\boxed{2}$

Geometric Progression:

a/ (1 - r) = 1/ (1 - 1/ 2) = 2