a snail starts moving towards a point 3 cm away at a pace of 1 cm per hour. as it gets tired , it covers only half the distance compared to previous hour in each succeeding hour . in how much time will the snail reach his target?

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$$\begin{gathered} \mathrm{The} \mathrm{snail} \mathrm{covers} 1 \mathrm{cm} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{hour}, \frac{1}{2} \mathrm{cm} \mathrm{in} \mathrm{the} \mathrm{next} \mathrm{hour}, \frac{1}{4} \mathrm{cm} \mathrm{in} \mathrm{the} \mathrm{next} \mathrm{hour} \mathrm{and} \mathrm{so} \mathrm{on}. \\ \mathrm{The} \mathrm{list} \mathrm{of} \mathrm{the} \mathrm{distance} \mathrm{covered}\left(\mathrm{in} \mathrm{cm}\right) \mathrm{by} \mathrm{snail} \mathrm{in} \mathrm{each} \mathrm{hour} \mathrm{is} : \\ 1, \frac{1}{2}, \frac{1}{4}, ......... \\ \mathrm{The} \mathrm{above} \mathrm{list} \mathrm{is} \mathrm{in} \mathrm{GP} \mathrm{with} \mathrm{a} = 1 \mathrm{and} \mathrm{r} = \frac{1}{2}. \\ \mathrm{Now}, \mathrm{Let} \mathrm{after} \mathrm{n} \mathrm{hours} \mathrm{snail} \mathrm{reaches} \mathrm{target}. \\ \mathrm{So}, {\mathrm{S}}_{\mathrm{n}} = 3 \\ \Rightarrow \frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}} = 3 \\ \Rightarrow \frac{1\left[1 - {\left(1/2\right)}^{\mathrm{n}}\right]}{1-1/2} = 3 \\ \Rightarrow 1 - {\left(\frac{1}{2}\right)}^{\mathrm{n}} = \frac{3}{2} \\ \Rightarrow {\left(\frac{1}{2}\right)}^{\mathrm{n}} = -\frac{1}{2} \\ \mathrm{Now}, \mathrm{there} \mathrm{is} \mathrm{no} \mathrm{value} \mathrm{of} \mathrm{n} \mathrm{such} \mathrm{that} \mathrm{above} \mathrm{equation} \mathrm{is} \mathrm{satisfied}. \\ \mathrm{So}, \mathrm{snail} \mathrm{will} \mathrm{not} \mathrm{reach} \mathrm{the} \mathrm{target} \mathrm{never}. \end{gathered}$$

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