Insert 3 numbers between 1 and 256 so that the resulting sequence is a GP

To insert three numbers between \(1\) and \(256\) to form a geometric progression (GP), the key is to find a common ratio, \(r\), such that the entire sequence becomes a GP. The sequence will look like:

\[
1, a_1, a_2, a_3, 256
\]

### Steps:

1. The general formula for a GP is:
   \[
   a_{n} = a_1 \cdot r^{n-1}
   \]
   where \(a_1\) is the first term and \(r\) is the common ratio.

2. For the sequence \(1, a_1, a_2, a_3, 256\), the common ratio \(r\) is determined as:
   \[
   r = \sqrt[4]{\frac{256}{1}} = \sqrt[4]{256} = 4
   \]

3. Calculate the intermediate terms:
   - \(a_1 = 1 \cdot 4 = 4\)
   - \(a_2 = 4 \cdot 4 = 16\)
   - \(a_3 = 16 \cdot 4 = 64\)

Thus, the sequence becomes:
\[
1, 4, 16, 64, 256
\]

### Final Answer:
The three numbers inserted are \(4\), \(16\), and \(64\).

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Would you like a detailed explanation of the steps or further examples? Here are five related questions to consider:

1. What is the general method to solve for terms in a GP with different start and end values?
2. How can we determine the common ratio when more than three terms need to be inserted?
3. What happens if the terms in the GP are non-integer values?
4. Can we derive the terms if the sequence is defined in reverse order?
5. How can the concept of a GP be applied in real-world problems like compound interest or population growth?

**Tip:** To ensure a sequence is a GP, check that the ratio between consecutive terms is consistent throughout the sequence.

Discover how to insert three numbers between 1 and 256 to form a geometric progression. Learn the step-by-step solution using the GP formula and properties of sequences, ideal for Grades 8-10 students.

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