Find the `5^(th)` term of the G.P. `(5)/(2),1`. . . . . . . . .

Answer

To find the 5th term of the geometric progression (G.P.) given as \( \frac{5}{2}, 1, \ldots \), we can follow these steps:

### Step 1: Identify the first term and the common ratio
The first term \( a \) of the G.P. is given as:
\[ a = \frac{5}{2} \]

To find the common ratio \( r \), we can divide the second term by the first term:
\[ r = \frac{1}{\frac{5}{2}} = \frac{1 \times 2}{5} = \frac{2}{5} \]

### Step 2: Use the formula for the nth term of a G.P.
The formula for the nth term \( T_n \) of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
For the 5th term, we set \( n = 5 \):
\[ T_5 = a \cdot r^{5-1} = a \cdot r^4 \]

### Step 3: Substitute the values of \( a \) and \( r \)
Now, substituting the values we found:
\[ T_5 = \frac{5}{2} \cdot \left(\frac{2}{5}\right)^4 \]

### Step 4: Calculate \( \left(\frac{2}{5}\right)^4 \)
Calculating \( \left(\frac{2}{5}\right)^4 \):
\[ \left(\frac{2}{5}\right)^4 = \frac{2^4}{5^4} = \frac{16}{625} \]

### Step 5: Substitute back to find \( T_5 \)
Now substituting back into the equation for \( T_5 \):
\[ T_5 = \frac{5}{2} \cdot \frac{16}{625} \]

### Step 6: Simplify the expression
Now, simplifying:
\[ T_5 = \frac{5 \cdot 16}{2 \cdot 625} = \frac{80}{1250} \]

### Step 7: Reduce the fraction
Now we can reduce \( \frac{80}{1250} \):
\[ \frac{80 \div 80}{1250 \div 80} = \frac{1}{15.625} = \frac{8}{125} \]

### Final Answer
Thus, the 5th term of the G.P. is:
\[ \boxed{\frac{8}{125}} \]