If a,b,c and d are four unequal positive numbers which are in A.P then

Options

A. `1/a+1/d gt 1/b+ 1/c`
B. `1/a+1/d lt 1/b+1/c`
C. `1/b+1/c gt 4/(a+d)`
D. `1/a+1/d=1/b+1/c`

Explanation

To solve the problem, we need to analyze the given conditions and derive the required relationships among the numbers \( a, b, c, \) and \( d \) which are in Arithmetic Progression (A.P).

### Step-by-Step Solution:

1. **Understanding A.P**:
Since \( a, b, c, d \) are in A.P, we can express them in terms of a common difference \( R \):
- \( b = a + R \)
- \( c = a + 2R \)
- \( d = a + 3R \)

2. **Finding \( \frac{1}{a} + \frac{1}{d} \) and \( \frac{1}{b} + \frac{1}{c} \)**:
We need to compare \( \frac{1}{a} + \frac{1}{d} \) with \( \frac{1}{b} + \frac{1}{c} \).

- Calculate \( \frac{1}{a} + \frac{1}{d} \):
\[ \frac{1}{a} + \frac{1}{d} = \frac{1}{a} + \frac{1}{a + 3R} = \frac{(a + 3R) + a}{a(a + 3R)} = \frac{2a + 3R}{a(a + 3R)} \]

- Calculate \( \frac{1}{b} + \frac{1}{c} \):
\[ \frac{1}{b} + \frac{1}{c} = \frac{1}{a + R} + \frac{1}{a + 2R} = \frac{(a + 2R) + (a + R)}{(a + R)(a + 2R)} = \frac{2a + 3R}{(a + R)(a + 2R)} \]

3. **Comparing the Two Expressions**:
Now we have:
\[ \frac{1}{a} + \frac{1}{d} = \frac{2a + 3R}{a(a + 3R)} \]
\[ \frac{1}{b} + \frac{1}{c} = \frac{2a + 3R}{(a + R)(a + 2R)} \]

To compare these two fractions, we can cross-multiply:
\[ (2a + 3R)(a + R)(a + 2R) \quad \text{and} \quad (2a + 3R)(a)(a + 3R) \]

Since \( 2a + 3R \) is common in both, we can simplify our comparison to:
\[ (a + R)(a + 2R) \quad \text{and} \quad a(a + 3R) \]

4. **Expanding Both Sides**:
- Left Side:
\[ (a + R)(a + 2R) = a^2 + 2aR + aR + 2R^2 = a^2 + 3aR + 2R^2 \]
- Right Side:
\[ a(a + 3R) = a^2 + 3aR \]

5. **Conclusion**:
From the above expansions, we see that:
\[ a^2 + 3aR + 2R^2 > a^2 + 3aR \implies 2R^2 > 0 \]
Since \( R \) is a positive number (as \( a, b, c, d \) are positive), this inequality holds true.

Therefore, we conclude:
\[ \frac{1}{a} + \frac{1}{d} > \frac{1}{b} + \frac{1}{c} \]

### Final Answer:
The correct statement is:
\[ \frac{1}{a} + \frac{1}{d} > \frac{1}{b} + \frac{1}{c} \]