To insert 3 numbers between 1 and 256 into an arithmetic progression, we need to calculate the common difference $d$ and identify the numbers that evenly space out the sequence.

Step-by-Step Solution:

1. The sequence should contain 1, 256, and three additional numbers.
2. The total sequence will have $n = 5$ terms.
3. In an arithmetic progression, the general term is given by: $a_n = a + (n-1)d$ where $a$ is the first term and $d$ is the common difference.
4. The 5th term of the sequence is 256, and the first term is 1: $256 = 1 + (5-1)d$ $256 = 1 + 4d$ $4d = 255 \quad \Rightarrow \quad d = 63.75$

Sequence:

• Starting with 1, add $d = 63.75$ repeatedly:
  1. First term: $1$
  2. Second term: $1 + 63.75 = 64.75$
  3. Third term: $64.75 + 63.75 = 128.5$
  4. Fourth term: $128.5 + 63.75 = 192.25$
  5. Fifth term: $192.25 + 63.75 = 256$

Resulting Sequence:

$1, 64.75, 128.5, 192.25, 256$

If you require integers, rounding the intermediate numbers can produce a slightly modified sequence: $1, 65, 129, 193, 256$

Would you like further clarification or a different type of sequence?

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Related Questions:

1. How do we calculate terms in a geometric progression instead of arithmetic progression?
2. What happens if we insert more or fewer numbers into the sequence?
3. How can we ensure the sequence contains only integers?
4. What is the formula for finding the sum of an arithmetic progression?
5. How would this process differ if the sequence was decreasing?

Tip: When working with arithmetic sequences, always check if the common difference $d$ is a rational number to simplify calculations.