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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?

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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!

Given the recurrence relation a_n = (2a_{n-1} - 1) / a_{n-1} and a_1 = 2, find a_4.

Learn how to solve a recurrence relation problem where a_n = (2a_{n-1} - 1) / a_{n-1} and a_1 = 2. This solution provides step-by-step calculations for a_2, a_3, and a_4. Ideal for high school students in grades 10-12.

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