The given problem asks to find $a_4$ from a recurrence relation:

$a_n = \frac{2a_{n-1} - 1}{a_{n-1}}$ with the initial condition $a_1 = 2$.

Let's calculate $a_2$, $a_3$, and $a_4$ step by step:

Step 1: Calculate $a_2$

Substitute $a_1 = 2$ into the recurrence relation: $a_2 = \frac{2a_1 - 1}{a_1} = \frac{2(2) - 1}{2} = \frac{4 - 1}{2} = \frac{3}{2}$

Step 2: Calculate $a_3$

Now, substitute $a_2 = \frac{3}{2}$ into the recurrence relation: $a_3 = \frac{2a_2 - 1}{a_2} = \frac{2\left(\frac{3}{2}\right) - 1}{\frac{3}{2}} = \frac{3 - 1}{\frac{3}{2}} = \frac{2}{\frac{3}{2}} = \frac{2 \times 2}{3} = \frac{4}{3}$

Step 3: Calculate $a_4$

Substitute $a_3 = \frac{4}{3}$ into the recurrence relation: $a_4 = \frac{2a_3 - 1}{a_3} = \frac{2\left(\frac{4}{3}\right) - 1}{\frac{4}{3}} = \frac{\frac{8}{3} - 1}{\frac{4}{3}} = \frac{\frac{8}{3} - \frac{3}{3}}{\frac{4}{3}} = \frac{\frac{5}{3}}{\frac{4}{3}} = \frac{5}{4}$

Thus, the value of $a_4$ is:

$a_4 = \frac{5}{4}$

Would you like more details on any of these steps, or have any questions?

Here are 5 related questions for further exploration:

1. What is the general formula for a recurrence relation?
2. How can you find the explicit formula for a recurrence relation like this one?
3. Can we solve this recurrence relation using generating functions?
4. What happens to the sequence $a_n$ as $n \to \infty$?
5. How can this recurrence relation be related to linear recurrence relations?

Tip: When working with recurrence relations, always carefully check the initial conditions—they are essential for building the sequence!