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. `q^2 = pr`
C. `p^2 = qr`
D. `pqr + pq + 1 = 0`

Answer

To solve the problem, we need to establish the relationship between the fourth, seventh, and tenth terms of a geometric progression (G.P.) given as \( p, q, r \) respectively.

### Step-by-step Solution:

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

2. **Express the Given Terms Using the General Formula**:
- For the 4th term:
\[ T_4 = a \cdot r^{4-1} = a \cdot r^3 = p \quad \text{(Equation 1)} \]
- For the 7th term:
\[ T_7 = a \cdot r^{7-1} = a \cdot r^6 = q \quad \text{(Equation 2)} \]
- For the 10th term:
\[ T_{10} = a \cdot r^{10-1} = a \cdot r^9 = r \quad \text{(Equation 3)} \]

3. **Manipulate the Equations**:
- From Equation 2, we can express \( a \) in terms of \( q \) and \( r \):
\[ a = \frac{q}{r^6} \quad \text{(from Equation 2)} \]
- Substitute \( a \) into Equation 1:
\[ \frac{q}{r^6} \cdot r^3 = p \]
Simplifying this gives:
\[ \frac{q \cdot r^3}{r^6} = p \implies \frac{q}{r^3} = p \implies q = p \cdot r^3 \quad \text{(Equation 4)} \]

4. **Substitute \( a \) into Equation 3**:
- Using \( a \) from Equation 2 in Equation 3:
\[ \frac{q}{r^6} \cdot r^9 = r \]
Simplifying this gives:
\[ \frac{q \cdot r^9}{r^6} = r \implies q \cdot r^3 = r \implies q = \frac{r}{r^3} \quad \text{(Equation 5)} \]

5. **Establish the Relationship**:
- Now we can multiply Equation 2 by itself:
\[ (a \cdot r^6)^2 = q^2 \]
This gives:
\[ a^2 \cdot r^{12} = q^2 \]
- From Equation 1 and Equation 3, we can express \( a^2 \):
\[ a^2 \cdot r^{12} = p \cdot r \]
- Therefore, we have:
\[ q^2 = p \cdot r \]

### Conclusion:
Thus, the relationship between the terms is:
\[ q^2 = p \cdot r \]