The first two terms of a G.P. are 125 and 25 respectively. Find the `5^(th)` and the `6^(th)` terms of the G.P.

Answer

To solve the problem, we need to find the 5th and 6th terms of a geometric progression (G.P.) where the first two terms are given as 125 and 25.

### Step 1: Identify the first term and the second term
Let the first term \( t_1 = 125 \) and the second term \( t_2 = 25 \).

### Step 2: Calculate the common ratio
The common ratio \( r \) of a G.P. can be found using the formula:
\[ r = \frac{t_2}{t_1} \]
Substituting the values:
\[ r = \frac{25}{125} = \frac{1}{5} \]

### Step 3: Use the formula for the nth term of a G.P.
The nth term of a G.P. is given by the formula:
\[ t_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 4: Find the 5th term
To find the 5th term \( t_5 \):
\[ t_5 = a \cdot r^{5-1} = 125 \cdot \left(\frac{1}{5}\right)^4 \]
Calculating \( \left(\frac{1}{5}\right)^4 \):
\[ \left(\frac{1}{5}\right)^4 = \frac{1}{625} \]
Now, substituting back:
\[ t_5 = 125 \cdot \frac{1}{625} = \frac{125}{625} = \frac{1}{5} \]

### Step 5: Find the 6th term
To find the 6th term \( t_6 \):
\[ t_6 = a \cdot r^{6-1} = 125 \cdot \left(\frac{1}{5}\right)^5 \]
Calculating \( \left(\frac{1}{5}\right)^5 \):
\[ \left(\frac{1}{5}\right)^5 = \frac{1}{3125} \]
Now, substituting back:
\[ t_6 = 125 \cdot \frac{1}{3125} = \frac{125}{3125} = \frac{1}{25} \]

### Final Answer
Thus, the 5th term \( t_5 \) is \( \frac{1}{5} \) and the 6th term \( t_6 \) is \( \frac{1}{25} \).

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