The fourth, seventh and tenth terms of a G.P. are p,q,r respectively, then

Options

A. `p^(2)=q^(2)+r^(2)`
B. `p^(2)=qr`
C. `q^(2)=pr`
D. `r^(2)=p^(2)+q^(2)`

Answer

To solve the problem, we need to find the relationship between the terms \( p, q, \) and \( r \) given that they are the 4th, 7th, and 10th terms of a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Identify the General Formula for Terms in a G.P.**:
The \( n \)-th term of a G.P. can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Write the Terms Based on Given Information**:
- The 4th term \( T_4 \) is given by:
\[ T_4 = a \cdot r^{4-1} = a \cdot r^3 = p \]
- The 7th term \( T_7 \) is given by:
\[ T_7 = a \cdot r^{7-1} = a \cdot r^6 = q \]
- The 10th term \( T_{10} \) is given by:
\[ T_{10} = a \cdot r^{10-1} = a \cdot r^9 = r \]

3. **Set Up Equations**:
From the above, we can write three equations:
- Equation 1: \( a \cdot r^3 = p \)
- Equation 2: \( a \cdot r^6 = q \)
- Equation 3: \( a \cdot r^9 = r \)

4. **Multiply Equation 1 and Equation 3**:
We multiply Equation 1 and Equation 3:
\[ p \cdot r = (a \cdot r^3) \cdot (a \cdot r^9) = a^2 \cdot r^{3+9} = a^2 \cdot r^{12} \]
Thus, we have:
\[ p \cdot r = a^2 \cdot r^{12} \quad \text{(Equation 4)} \]

5. **Square Equation 2**:
Now, we square Equation 2:
\[ q^2 = (a \cdot r^6)^2 = a^2 \cdot r^{12} \quad \text{(Equation 5)} \]

6. **Relate Equations 4 and 5**:
From Equation 4 and Equation 5, we can equate:
\[ p \cdot r = q^2 \]

7. **Final Relationship**:
Therefore, we conclude that:
\[ p \cdot r = q^2 \]

### Final Answer:
The relationship between \( p, q, \) and \( r \) is:
\[ p \cdot r = q^2 \]