Find the 20th and nth term of the G.P. `(5)/(2),(5)/(4),(5)/(8),…`

Explanation

To find the 20th and nth term of the given geometric progression (G.P.) \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots \), we can follow these steps:

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

To find the common ratio \( r \), we can divide the second term by the first term:
\[ r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{5}{4} \times \frac{2}{5} = \frac{2}{4} = \frac{1}{2} \]

### Step 2: Find the 20th term
The formula for the \( n \)-th term of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
To find the 20th term, we substitute \( n = 20 \):
\[ T_{20} = a \cdot r^{20-1} = \frac{5}{2} \cdot \left(\frac{1}{2}\right)^{19} \]
Calculating this gives:
\[ T_{20} = \frac{5}{2} \cdot \frac{1}{2^{19}} = \frac{5}{2^{20}} \]

### Step 3: Find the nth term
Using the same formula for the \( n \)-th term:
\[ T_n = a \cdot r^{n-1} = \frac{5}{2} \cdot \left(\frac{1}{2}\right)^{n-1} \]
This simplifies to:
\[ T_n = \frac{5}{2} \cdot \frac{1}{2^{n-1}} = \frac{5}{2^n} \]

### Final Results
Thus, the 20th term of the G.P. is:
\[ T_{20} = \frac{5}{2^{20}} \]
And the \( n \)-th term of the G.P. is:
\[ T_n = \frac{5}{2^n} \]