Insert 5 geometric means between `(1)/(3)` and 243.

Answer

To insert 5 geometric means between \( \frac{1}{3} \) and 243, we will follow these steps:

### Step 1: Identify the first and last terms
The first term \( a \) is \( \frac{1}{3} \) and the last term (7th term) \( a_7 \) is 243.

### Step 2: Use the formula for the nth term of a geometric progression
The nth term of a geometric progression is given by:
\[ a_n = a \cdot r^{n-1} \]
Here, \( a_7 = a \cdot r^{6} \) (since we have 5 terms in between, making it the 7th term).

### Step 3: Set up the equation
Substituting the known values into the equation:
\[ 243 = \frac{1}{3} \cdot r^{6} \]

### Step 4: Solve for \( r^6 \)
To eliminate \( \frac{1}{3} \), multiply both sides by 3:
\[ 3 \cdot 243 = r^{6} \]
\[ 729 = r^{6} \]

### Step 5: Find \( r \)
Now, take the sixth root of both sides:
\[ r = 729^{\frac{1}{6}} \]
Since \( 729 = 3^6 \), we have:
\[ r = 3 \]

### Step 6: Calculate the geometric means
Now that we have \( r \), we can find the 5 geometric means:
1. First term: \( a = \frac{1}{3} \)
2. Second term: \( a \cdot r = \frac{1}{3} \cdot 3 = 1 \)
3. Third term: \( a \cdot r^2 = \frac{1}{3} \cdot 3^2 = \frac{1}{3} \cdot 9 = 3 \)
4. Fourth term: \( a \cdot r^3 = \frac{1}{3} \cdot 3^3 = \frac{1}{3} \cdot 27 = 9 \)
5. Fifth term: \( a \cdot r^4 = \frac{1}{3} \cdot 3^4 = \frac{1}{3} \cdot 81 = 27 \)
6. Sixth term: \( a \cdot r^5 = \frac{1}{3} \cdot 3^5 = \frac{1}{3} \cdot 243 = 81 \)

### Step 7: List the geometric means
The 5 geometric means between \( \frac{1}{3} \) and 243 are:
1. \( 1 \)
2. \( 3 \)
3. \( 9 \)
4. \( 27 \)
5. \( 81 \)

### Final Answer
The 5 geometric means between \( \frac{1}{3} \) and 243 are \( 1, 3, 9, 27, \) and \( 81 \).

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