If `a+c`, `a+b`, `b+c` are in `G.P` and `a,c,b` are in `H.P.` where `a`,`b`,`c gt 0`, then the value of `(a+b)/(c )` is

Options

A. `3`
B. `2`
C. `(3)/(2)`
D. `4`

Explanation

To solve the problem, we need to find the value of \(\frac{a+b}{c}\) given that \(a+c\), \(a+b\), and \(b+c\) are in geometric progression (G.P.) and \(a\), \(b\), and \(c\) are in harmonic progression (H.P.).

### Step-by-step Solution:

1. **Understanding G.P. Condition**:
Since \(a+c\), \(a+b\), and \(b+c\) are in G.P., we can use the property of G.P. which states that the square of the middle term is equal to the product of the other two terms:
\[ (a+b)^2 = (a+c)(b+c) \]

2. **Expanding the G.P. Equation**:
Expanding both sides:
\[ (a+b)^2 = a^2 + 2ab + b^2 \]
\[ (a+c)(b+c) = ab + ac + bc + c^2 \]
So we have:
\[ a^2 + 2ab + b^2 = ab + ac + bc + c^2 \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ a^2 + ab + b^2 - ac - bc - c^2 = 0 \]

4. **Understanding H.P. Condition**:
Since \(a\), \(b\), and \(c\) are in H.P., we can express this condition as:
\[ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in A.P.} \]
This implies:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]
Multiplying through by \(abc\) gives:
\[ 2ac = ab + bc \]

5. **Expressing \(ab\)**:
From the H.P. condition, we can express \(ab\) as:
\[ ab = 2ac - bc \]

6. **Substituting \(ab\) into the G.P. Equation**:
Substitute \(ab\) back into the rearranged G.P. equation:
\[ a^2 + (2ac - bc) + b^2 - ac - bc - c^2 = 0 \]
Simplifying this leads to:
\[ a^2 + b^2 + ac - 2bc - c^2 = 0 \]

7. **Finding \(a+b\)**:
We can express \(a+b\) in terms of \(c\):
\[ a + b = 2c \]

8. **Calculating \(\frac{a+b}{c}\)**:
Now we can find:
\[ \frac{a+b}{c} = \frac{2c}{c} = 2 \]

### Final Answer:
Thus, the value of \(\frac{a+b}{c}\) is \(2\).