Four geometric means are inserted between 1/8 and 128. Find the third geometric mean.

Options

A. 4
B. 16
C. 32
D. 8

Answer

To find the third geometric mean inserted between \( \frac{1}{8} \) and \( 128 \), we can follow these steps:

### Step 1: Understand the geometric progression
In a geometric progression (GP), if \( a_1 \) is the first term and \( a_n \) is the last term, the terms can be expressed as:
- \( a_1 = \frac{1}{8} \)
- \( a_2 = x_1 \) (first geometric mean)
- \( a_3 = x_2 \) (second geometric mean)
- \( a_4 = x_3 \) (third geometric mean)
- \( a_5 = x_4 \) (fourth geometric mean)
- \( a_6 = 128 \)

### Step 2: Use the formula for the nth term of a GP
The nth term of a GP can be expressed as:
\[ a_n = a_1 \cdot r^{(n-1)} \]
where \( r \) is the common ratio.

### Step 3: Set up the equation for the last term
For our case, we have:
\[ 128 = \frac{1}{8} \cdot r^5 \]

### Step 4: Solve for \( r \)
To solve for \( r \), we first multiply both sides by \( 8 \):
\[ 128 \cdot 8 = r^5 \]
\[ 1024 = r^5 \]

Now, take the fifth root of both sides:
\[ r = 1024^{\frac{1}{5}} \]
Calculating \( 1024^{\frac{1}{5}} \):
\[ 1024 = 2^{10} \quad \text{so} \quad 1024^{\frac{1}{5}} = 2^{10/5} = 2^2 = 4 \]
Thus, \( r = 4 \).

### Step 5: Find the geometric means
Now that we have \( r \), we can calculate the geometric means:
- \( a_2 = a_1 \cdot r = \frac{1}{8} \cdot 4 = \frac{4}{8} = \frac{1}{2} \)
- \( a_3 = a_2 \cdot r = \frac{1}{2} \cdot 4 = 2 \) (this is the third geometric mean)
- \( a_4 = a_3 \cdot r = 2 \cdot 4 = 8 \)
- \( a_5 = a_4 \cdot r = 8 \cdot 4 = 32 \)

### Step 6: Verify the last term
Finally, we check the last term:
\[ a_6 = a_5 \cdot r = 32 \cdot 4 = 128 \]
This confirms our calculations are correct.

### Conclusion
The third geometric mean is \( 2 \).

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