Question

Question 127(i)

An insect is on the 0 point of a number line, hopping towards 1. It covers half the distance from its current location to 1 with each hop.
So, it will be at $\frac{1}{2}$ after one hop, $\frac{3}{4}$ after two hops and so on.



Make a table showing the insects location for the first 10 hops.

Answer 1

On the basis of given information in the question, we can arrange the following table which shows the insect's location for the first 10 hops.

$\begin{array}{cccc}Numberofhops& Distancecovered& Distanceleft& Distancecovered\\ 1.& \frac{1}{2}& \frac{1}{2}& 1-\frac{1}{2}\\ 2.& \frac{1}{2}\left(\frac{1}{2}\right)+\frac{1}{2}& \frac{1}{4}& 1-\frac{1}{4}\\ 3.& \frac{1}{2}\left(\frac{1}{4}\right)+\frac{3}{4}& \frac{1}{8}& 1-\frac{1}{8}\\ 4.& \frac{1}{2}\left(\frac{1}{8}\right)+\frac{7}{8}& \frac{1}{16}& 1-\frac{1}{16}\\ 5.& \frac{1}{2}\left(\frac{1}{16}\right)+\frac{15}{16}& \frac{1}{32}& 1-\frac{1}{32}\\ 6.& \frac{1}{2}\left(\frac{1}{32}\right)+\frac{31}{32}& \frac{1}{64}& 1-\frac{1}{64}\\ 7.& \frac{1}{2}\left(\frac{1}{64}\right)+\frac{63}{64}& \frac{1}{128}& 1-\frac{1}{128}\\ 8.& \frac{1}{2}\left(\frac{1}{128}\right)+\frac{127}{128}& \frac{1}{256}& 1-\frac{1}{256}\\ 9.& \frac{1}{2}\left(\frac{1}{256}\right)+\frac{255}{256}& \frac{1}{512}& 1-\frac{1}{512}\\ 10.& \frac{1}{2}\left(\frac{1}{512}\right)+\frac{511}{512}& \frac{1}{1024}& 1-\frac{1}{1024}\end{array}$