For the G.P. `(1)/(27),(1)/(9),(1)/(3), . . . . . .. . . . .. ,81,`
find the product of fourth term from the beginning and the fourth term from the beginning and the fourth term from the end.

Answer

To solve the problem, we need to find the product of the fourth term from the beginning and the fourth term from the end of the given geometric progression (G.P.):

\[ \frac{1}{27}, \frac{1}{9}, \frac{1}{3}, \ldots, 81 \]

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the G.P. is given as:

\[ a = \frac{1}{27} \]

To find the common ratio \( r \), we can take the second term and divide it by the first term:

\[ r = \frac{\frac{1}{9}}{\frac{1}{27}} = \frac{1}{9} \times \frac{27}{1} = 3 \]

### Step 2: Identify the last term
The last term \( L \) of the G.P. is given as:

\[ L = 81 \]

### Step 3: Find the fourth term from the beginning
The formula for the \( n \)-th term of a G.P. is given by:

\[ T_n = a \cdot r^{n-1} \]

To find the fourth term from the beginning (\( T_4 \)):

\[ T_4 = a \cdot r^{4-1} = \frac{1}{27} \cdot 3^3 \]

Calculating \( 3^3 \):

\[ 3^3 = 27 \]

Now substituting back:

\[ T_4 = \frac{1}{27} \cdot 27 = 1 \]

### Step 4: Find the fourth term from the end
To find the fourth term from the end, we can use the formula for the \( n \)-th term from the end, which is:

\[ T_{n-k+1} = L \cdot \frac{1}{r^{k-1}} \]

Here, \( k = 4 \) (since we want the fourth term from the end):

\[ T_{n-3} = 81 \cdot \frac{1}{3^{4-1}} = 81 \cdot \frac{1}{3^3} \]

Calculating \( 3^3 \):

\[ 3^3 = 27 \]

Now substituting back:

\[ T_{n-3} = 81 \cdot \frac{1}{27} = 3 \]

### Step 5: Find the product of the two terms
Now, we need to find the product of the fourth term from the beginning and the fourth term from the end:

\[ \text{Product} = T_4 \cdot T_{n-3} = 1 \cdot 3 = 3 \]

### Final Answer
Thus, the product of the fourth term from the beginning and the fourth term from the end is:

\[ \boxed{3} \]