To insert 3 numbers between 1 and 256 so that the sequence becomes arithmetic, the numbers should divide the total difference between 1 and 256 evenly. An arithmetic sequence has a constant difference $d$ between consecutive terms.

Steps:

1. The sequence starts with 1 and ends with 256.

2. There will be a total of 5 terms after inserting 3 numbers (1, $a_1$, $a_2$, $a_3$, 256).

3. The common difference $d$ is calculated as: $d = \frac{\text{Last term} - \text{First term}}{\text{Number of gaps}}$ Substituting: $d = \frac{256 - 1}{4} = 63.75$

4. Now, the sequence becomes: $1, 1 + 63.75, 1 + 2 \cdot 63.75, 1 + 3 \cdot 63.75, 256$

5. The inserted numbers are: $64.75, 128.5, 192.25$

Final Sequence:

$1, 64.75, 128.5, 192.25, 256$

Would you like me to explain this further or provide another example?

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Here are 5 related questions to explore:

1. What are the steps to generalize arithmetic sequences with more inserted terms?
2. How can we adjust the sequence if the inserted numbers must be integers?
3. Can you derive the formula for any number of terms in an arithmetic sequence?
4. What happens if we insert numbers to form a geometric sequence instead?
5. How does this method change if the sequence must start and end with different numbers?

Tip: Arithmetic sequences are evenly spaced. When solving, focus on evenly dividing the interval.