2B. Given are the lines $y=\frac{3}{4} x+6, y=\frac{3}{4} x+3$ and the point $T(7,24)$. Determine the equation of the line that contains the given point, and the segment with endpoints obtained by the intersection of this line with the given lines, has a length of 4.
Solution. The equation of the desired line is $y-24=k(x-7)$. At its intersection, the points $A\left(\frac{7 k-18}{k-\frac{3}{4}}, \frac{\frac{45}{4} k-18}{k-\frac{3}{4}}\right)$ and $B\left(\frac{7 k-21}{k-\frac{3}{4}}, \frac{\frac{33}{4} k-18}{k-\frac{3}{4}}\right)$ are obtained.

From the condition of the problem, we get $\left(\frac{3}{k-\frac{3}{4}}\right)^{2}+\left(\frac{3 k}{k-\frac{3}{4}}\right)^{2}=16$, which simplifies to the quadratic equation $7 k^{2}-24 k=0$. Its solutions are $k_{1}=0$ and $k_{2}=\frac{24}{7}$. Therefore, the desired lines are $y=24$ and $y=\frac{24}{7} x$.

3AB. A sequence of positive real numbers $a_{0}, a_{1}, a_{2}, \ldots$ is given such that

$$
a_{1}=1-a_{0}, a_{n+1}=1-a_{n}\left(1-a_{n}\right) \text {, for } n \geq 1 \text {. }
$$

Prove that for every natural number $n$,

$$
a_{0} a_{1} \cdot \ldots \cdot a_{n}\left(\frac{1}{a_{0}}+\frac{1}{a_{1}}+\ldots+\frac{1}{a_{n}}\right)=1
$$