Find the 5th term from the end of the G .P. `(1)/(512),(1)/(256),(1)/(128),...256.`

Explanation

To find the 5th term from the end of the given geometric progression (G.P.) \( \frac{1}{512}, \frac{1}{256}, \frac{1}{128}, \ldots, 256 \), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the G.P. is \( \frac{1}{512} \).

To find the common ratio \( r \), we can take the ratio of the second term to the first term:
\[ r = \frac{\frac{1}{256}}{\frac{1}{512}} = \frac{1}{256} \times \frac{512}{1} = \frac{512}{256} = 2 \]

### Step 2: Identify the last term
The last term \( L \) of the G.P. is given as \( 256 \).

### Step 3: Use the formula for the nth term from the end of a G.P.
The formula for the \( n \)-th term from the end of a G.P. is given by:
\[ T_n = \frac{L}{r^{n-1}} \]
where \( L \) is the last term, \( r \) is the common ratio, and \( n \) is the term number from the end.

### Step 4: Substitute the values to find the 5th term from the end
We need to find the 5th term from the end, so we set \( n = 5 \):
\[ T_5 = \frac{256}{2^{5-1}} = \frac{256}{2^4} \]

### Step 5: Simplify the expression
Calculating \( 2^4 \):
\[ 2^4 = 16 \]
Now substitute this back into the equation:
\[ T_5 = \frac{256}{16} \]

### Step 6: Perform the division
Calculating \( \frac{256}{16} \):
\[ T_5 = 16 \]

Thus, the 5th term from the end of the G.P. is \( \boxed{16} \).

---