If three geometric means inserted between 1 and 256 , then find the common ratio

Answer

To solve the problem of finding the common ratio when three geometric means are inserted between 1 and 256, we can follow these steps:

### Step 1: Understand the Problem
We are given two numbers, 1 and 256, and we need to insert three geometric means (let's call them \( g_1, g_2, g_3 \)) between them. This means we have the sequence:
\[ 1, g_1, g_2, g_3, 256 \]

### Step 2: Identify the Terms of the Geometric Progression (GP)
In a geometric progression, the \( n \)-th term can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

Here, the first term \( a = 1 \) and the fifth term (which is 256) can be represented as:
\[ a_5 = 1 \cdot r^{5-1} = r^4 \]

### Step 3: Set Up the Equation
Since the fifth term is given as 256, we can set up the equation:
\[ r^4 = 256 \]

### Step 4: Solve for the Common Ratio \( r \)
To find \( r \), we need to take the fourth root of both sides:
\[ r = \sqrt[4]{256} \]

### Step 5: Calculate the Fourth Root
We know that:
\[ 256 = 4^4 \]
Thus,
\[ r = \sqrt[4]{4^4} = 4 \]

### Step 6: Consider the Negative Root
Since the common ratio in a geometric progression can also be negative, we also consider:
\[ r = -4 \]

### Final Answer
Therefore, the common ratio can be:
\[ r = 4 \quad \text{or} \quad r = -4 \]

### Summary
The common ratio when three geometric means are inserted between 1 and 256 is \( r = \pm 4 \).

---