https://file.aiquickdraw.com/mathbot/file/1726730220291ragbrr8w.jpg

Indicate the property, definition, or theorem as needed.

We are given a recursive sequence with:

\[
a_1 = 1 \quad \text{and} \quad a_{n+1} = \frac{1}{2} \left( a_n + \frac{4}{a_n} \right).
\]

We are asked to compute \( a_5 \).

Let's calculate \( a_2, a_3, a_4, \) and \( a_5 \) step by step.

### Step-by-Step Calculation:

1. **For \( a_2 \):**
   \[
   a_2 = \frac{1}{2} \left( a_1 + \frac{4}{a_1} \right) = \frac{1}{2} \left( 1 + \frac{4}{1} \right) = \frac{1}{2} \times 5 = 2.5.
   \]

2. **For \( a_3 \):**
   \[
   a_3 = \frac{1}{2} \left( a_2 + \frac{4}{a_2} \right) = \frac{1}{2} \left( 2.5 + \frac{4}{2.5} \right) = \frac{1}{2} \left( 2.5 + 1.6 \right) = \frac{1}{2} \times 4.1 = 2.05.
   \]

3. **For \( a_4 \):**
   \[
   a_4 = \frac{1}{2} \left( a_3 + \frac{4}{a_3} \right) = \frac{1}{2} \left( 2.05 + \frac{4}{2.05} \right) = \frac{1}{2} \left( 2.05 + 1.9512 \right) = \frac{1}{2} \times 4.0012 = 2.0006.
   \]

4. **For \( a_5 \):**
   \[
   a_5 = \frac{1}{2} \left( a_4 + \frac{4}{a_4} \right) = \frac{1}{2} \left( 2.0006 + \frac{4}{2.0006} \right) = \frac{1}{2} \left( 2.0006 + 1.9994 \right) = \frac{1}{2} \times 4.0000 = 2.
   \]

### Conclusion:
Thus, \( a_5 = 2 \).

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What happens to \( a_n \) as \( n \) becomes large?
2. Is there a closed-form expression for \( a_n \)?
3. Can we prove that \( a_n \) converges, and if so, to what value?
4. How does this recursive sequence relate to the arithmetic-geometric mean?
5. How would the result change if the initial condition \( a_1 \) were different?

**Tip:** When working with recursive sequences, calculating a few terms manually can reveal patterns that may help predict long-term behavior.

Suppose a₁ = 1 and aₙ₊₁ = ½ (aₙ + 4/aₙ). Compute a₅.

Learn how to compute the fifth term, a₅, in a recursive sequence with the given formula aₙ₊₁ = ½ (aₙ + 4/aₙ) and initial condition a₁ = 1. This step-by-step solution is suitable for high school students and covers recursion and algebraic sequence behavior.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation