To travel a straight path of 10 meters in length, a flea uses the following strategy: each day, it covers half of the remaining distance. Thus, it travels 5 meters on the first day, 2.5 meters on the second, and so on (the size of the flea can be disregarded).

(a) How many meters will it have traveled by the end of the seventh day? And by the tenth?

(b) From which day will the flea be less than $0.001 \mathrm{~m}$ from the end of the path?
On the first jump, the flea covers $10 \times\left(\frac{1}{2}\right) \mathrm{m}$, on the second jump, it covers $10 \times\left(\frac{1}{2}\right)^{2} \mathrm{~m}$, and so on. After 7 jumps, the flea will have covered

$$
\begin{aligned}
10\left(\frac{1}{2}\right) & +10\left(\frac{1}{2}\right)^{2}+10\left(\frac{1}{2}\right)^{3}+10\left(\frac{1}{2}\right)^{4}+10\left(\frac{1}{2}\right)^{5}+10\left(\frac{1}{2}\right)^{6}+10\left(\frac{1}{2}\right)^{7} \\
& =10\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{5}+\left(\frac{1}{2}\right)^{6}+\left(\frac{1}{2}\right)^{7}\right] \\
& =10 \times \frac{2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2+1}{2^{7}}=10 \times \frac{127}{128} \approx 9.9
\end{aligned}
$$

Thus, in 7 days, it will have covered approximately $9.9 \mathrm{~m}$. In general, after $n$ days, the flea will have covered

$$
10\left(\frac{1}{2}+\cdots+\frac{1}{2^{n}}\right) \text { meters. }
$$

To calculate the sum above, note that $\frac{1}{2}+\cdots+\frac{1}{2^{n}}$ is the sum of the $n$ terms of a geometric progression whose first term is $a_{1}=1 / 2$ and whose common ratio is $q=1 / 2$. The formula for this sum is

$$
S_{n}=\frac{a_{1}\left(1-q^{n}\right)}{1-q}=\frac{1 / 2\left(1-1 / 2^{n}\right)}{1-1 / 2}=1-\frac{1}{2^{n}}
$$

Thus,

$$
10\left(\frac{1}{2}+\cdots+\frac{1}{2^{n}}\right)=10\left(1-\frac{1}{2^{n}}\right)=10-\frac{10}{2^{n}}
$$

Taking $n=10$, we get

$$
10-\frac{10}{2^{10}}=10-\frac{10}{1024}=10 \times \frac{1023}{1024} \approx 9.99
$$

Therefore, by the end of the tenth day, the flea will have covered approximately 9.99 meters.

The flea will be less than $0.001 \mathrm{~m}$ from the end of the path when it has already covered at least $9.999=10-0.001$ meters, that is, when

$$
10-\frac{10}{2^{n}} \geq 10-0.0001
$$

which is equivalent to $0.001 \geq 10 / 2^{n}$, or $2^{n} \geq 10 / 0.001=10000$.

Now, $2^{13}=2^{10} \times 2^{3}=1024 \times 8=8192<10000<16384=2^{14}$, so we must take $n=14$ and the flea will be less than $0.001 \mathrm{~m}$ from the end of the path starting from the fourteenth day.