If the `4^(th),7^(th)` and `10^(th)` terms of a G.P. are a, b and c respectively, prove that : `b^(2)=ac`

Answer

To prove that \( b^2 = ac \) where the 4th, 7th, and 10th terms of a geometric progression (G.P.) are \( a \), \( b \), and \( c \) respectively, we can follow these steps:

### Step-by-step Solution:

1. **Identify 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. **Write the equations for the given terms**:
- For the 4th term:
\[ T_4 = A \cdot R^{4-1} = A \cdot R^3 = a \quad \text{(1)} \]
- For the 7th term:
\[ T_7 = A \cdot R^{7-1} = A \cdot R^6 = b \quad \text{(2)} \]
- For the 10th term:
\[ T_{10} = A \cdot R^{10-1} = A \cdot R^9 = c \quad \text{(3)} \]

3. **Express \( A \) in terms of \( a \)**:
From equation (1):
\[ A = \frac{a}{R^3} \quad \text{(4)} \]

4. **Substitute \( A \) into equations (2) and (3)**:
- Substitute into equation (2):
\[ b = A \cdot R^6 = \frac{a}{R^3} \cdot R^6 = a \cdot R^3 \quad \text{(5)} \]
- Substitute into equation (3):
\[ c = A \cdot R^9 = \frac{a}{R^3} \cdot R^9 = a \cdot R^6 \quad \text{(6)} \]

5. **Multiply equations (5) and (6)**:
\[ b \cdot c = (a \cdot R^3) \cdot (a \cdot R^6) = a^2 \cdot R^{3+6} = a^2 \cdot R^9 \]

6. **Express \( b^2 \)**:
\[ b^2 = (a \cdot R^3)^2 = a^2 \cdot R^{6} \]

7. **Relate \( b^2 \) and \( ac \)**:
We have:
\[ ac = a \cdot (a \cdot R^6) = a^2 \cdot R^6 \]
Thus:
\[ b^2 = ac \]

### Conclusion:
We have shown that \( b^2 = ac \).