If 1, x, y, z, 16 are in geometric progression, then what is the value of x+y+z?

Options

A. 8
B. 12
C. 14
D. 16

Answer

To solve the problem where the numbers 1, x, y, z, and 16 are in geometric progression (GP), we can follow these steps:

### Step 1: Understand the properties of GP
In a geometric progression, the ratio of consecutive terms is constant. This means that if we denote the common ratio as \( r \), then we can express the terms as follows:
- \( x = 1 \cdot r \)
- \( y = x \cdot r = 1 \cdot r^2 \)
- \( z = y \cdot r = 1 \cdot r^3 \)
- \( 16 = z \cdot r = 1 \cdot r^4 \)

### Step 2: Set up the equation
From the last term, we can write:
\[ 16 = r^4 \]

### Step 3: Solve for \( r \)
To find \( r \), we take the fourth root of both sides:
\[ r = \sqrt[4]{16} = \sqrt[4]{2^4} = 2 \]

### Step 4: Find the values of \( x \), \( y \), and \( z \)
Now that we have \( r = 2 \), we can find the values of \( x \), \( y \), and \( z \):
- \( x = 1 \cdot r = 1 \cdot 2 = 2 \)
- \( y = 1 \cdot r^2 = 1 \cdot 2^2 = 4 \)
- \( z = 1 \cdot r^3 = 1 \cdot 2^3 = 8 \)

### Step 5: Calculate \( x + y + z \)
Now we can find the sum:
\[ x + y + z = 2 + 4 + 8 = 14 \]

### Final Answer
The value of \( x + y + z \) is \( 14 \).

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