The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that `q^(2)=ps.`

Explanation

To solve the problem, we need to show that \( q^2 = ps \) given that the 5th, 8th, and 11th terms of a geometric progression (G.P.) are \( p \), \( q \), and \( s \) respectively.

### Step-by-Step Solution:

1. **Define the Terms of the G.P.:**
In a geometric progression, the \( n \)-th term is given by:
\[ A_n = A \cdot r^{n-1} \]
where \( A \) is the first term and \( r \) is the common ratio.

2. **Express the Given Terms:**
- The 5th term (\( A_5 \)) is:
\[ A_5 = A \cdot r^{5-1} = A \cdot r^4 = p \]
- The 8th term (\( A_8 \)) is:
\[ A_8 = A \cdot r^{8-1} = A \cdot r^7 = q \]
- The 11th term (\( A_{11} \)) is:
\[ A_{11} = A \cdot r^{11-1} = A \cdot r^{10} = s \]

3. **Square the 8th Term:**
We need to find \( q^2 \):
\[ q^2 = (A \cdot r^7)^2 = A^2 \cdot r^{14} \]

4. **Express \( ps \):**
Now, we will express \( ps \):
\[ ps = (A \cdot r^4)(A \cdot r^{10}) = A^2 \cdot r^{4 + 10} = A^2 \cdot r^{14} \]

5. **Equate \( q^2 \) and \( ps \):**
From the previous steps, we have:
\[ q^2 = A^2 \cdot r^{14} \]
and
\[ ps = A^2 \cdot r^{14} \]
Therefore, we can conclude:
\[ q^2 = ps \]

### Conclusion:
We have shown that \( q^2 = ps \).