If a,b and c are three positive numbers in an arithmetic progression, then :

Options

A. `ac gt b^(2)`
B. `b^(2) gt a+c`
C. `ab+bc le2ac`
D. `ab+bcge2ac`

Answer

To solve the problem, we need to establish the relationship between three positive numbers \( a \), \( b \), and \( c \) that are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression**:
- If \( a \), \( b \), and \( c \) are in arithmetic progression, by definition, the middle term \( b \) is the average of the other two terms. This can be expressed mathematically as:
\[ b = \frac{a + c}{2} \]

2. **Reciprocal Relationship**:
- It is given that if \( a \), \( b \), and \( c \) are in AP, then their reciprocals \( \frac{1}{a} \), \( \frac{1}{b} \), and \( \frac{1}{c} \) are in harmonic progression (HP).

3. **Finding the Harmonic Mean**:
- The harmonic mean (HM) of \( a \), \( b \), and \( c \) can be expressed as:
\[ b = \frac{2ac}{a + c} \]

4. **Inequality Relation**:
- From the properties of means, we know that the arithmetic mean is always greater than or equal to the harmonic mean:
\[ \frac{a + c}{2} \geq \frac{2ac}{a + c} \]

5. **Cross Multiplication**:
- To eliminate the fractions, we can cross-multiply:
\[ (a + c)^2 \geq 4ac \]

6. **Expanding and Rearranging**:
- Expanding the left side gives:
\[ a^2 + 2ac + c^2 \geq 4ac \]
- Rearranging this inequality results in:
\[ a^2 - 2ac + c^2 \geq 0 \]

7. **Factoring**:
- The left-hand side can be factored as:
\[ (a - c)^2 \geq 0 \]
- This inequality holds true since the square of any real number is non-negative.

8. **Conclusion**:
- Therefore, we conclude that if \( a \), \( b \), and \( c \) are in arithmetic progression, then the relation \( ab + bc \geq 2ac \) holds true.

### Final Relation:
Thus, the required condition is:
\[ ab + bc \geq 2ac \]