Find the third term from the end of the G.P.
`(2)/(27),(2)/(9),(2)/(3), . . . . . . . . . . .. . . . .,162`.

Answer

To find the third term from the end of the geometric progression (G.P.) given by the terms \( \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162 \), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the G.P. is:
\[ a = \frac{2}{27} \]

To find the common ratio \( r \), we can take the second term and divide it by the first term:
\[ r = \frac{\frac{2}{9}}{\frac{2}{27}} = \frac{2}{9} \times \frac{27}{2} = \frac{27}{9} = 3 \]

### Step 2: Write the general term of the G.P.
The \( n \)-th term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
Substituting the values of \( a \) and \( r \):
\[ T_n = \frac{2}{27} \cdot 3^{n-1} \]

### Step 3: Find the number of terms in the G.P.
We know that the last term is \( 162 \). We can set up the equation:
\[ 162 = \frac{2}{27} \cdot 3^{n-1} \]
To solve for \( n \), we first multiply both sides by \( 27 \):
\[ 162 \cdot 27 = 2 \cdot 3^{n-1} \]
Calculating \( 162 \cdot 27 \):
\[ 162 \cdot 27 = 4374 \]
So we have:
\[ 4374 = 2 \cdot 3^{n-1} \]
Dividing both sides by \( 2 \):
\[ 2187 = 3^{n-1} \]
Now we can express \( 2187 \) as a power of \( 3 \):
\[ 2187 = 3^7 \]
Thus, we have:
\[ 3^{n-1} = 3^7 \implies n - 1 = 7 \implies n = 8 \]
This means there are 8 terms in the G.P.

### Step 4: Find the third term from the end
The third term from the end corresponds to the \( (n-2) \)-th term:
\[ T_{n-2} = T_{8-2} = T_6 \]
Using the formula for the \( n \)-th term:
\[ T_6 = \frac{2}{27} \cdot 3^{6-1} = \frac{2}{27} \cdot 3^5 \]
Calculating \( 3^5 \):
\[ 3^5 = 243 \]
Thus:
\[ T_6 = \frac{2}{27} \cdot 243 = \frac{2 \cdot 243}{27} = \frac{486}{27} = 18 \]

### Final Answer
The third term from the end of the G.P. is:
\[ \boxed{18} \]