If three positive real numbers a,b,c, `(cgta)` are in H.P., then `log(a+c)+log(a-2b+c)` is equal to

Options

A. 2 log (c-b)
B. 2 log (a+c)
C. 2 log (c-a)
D. log a+log b+log c

Answer

To solve the problem, we need to find the value of \( \log(a+c) + \log(a-2b+c) \) given that \( a, b, c \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding H.P.**:
If three numbers \( a, b, c \) are in H.P., then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.). This means:
\[ 2b = \frac{1}{a} + \frac{1}{c} \]
Rearranging gives:
\[ 2b = \frac{c + a}{ac} \implies b = \frac{ac}{a+c} \]

2. **Substituting for \( b \)**:
We need to substitute this value of \( b \) into the expression \( \log(a+c) + \log(a-2b+c) \):
\[ \log(a+c) + \log\left(a - 2\left(\frac{ac}{a+c}\right) + c\right) \]

3. **Simplifying \( a - 2b + c \)**:
Substitute \( b \):
\[ a - 2b + c = a - 2\left(\frac{ac}{a+c}\right) + c \]
Finding a common denominator:
\[ = \frac{(a+c)(a) - 2ac + (a+c)(c)}{a+c} \]
Simplifying the numerator:
\[ = \frac{a^2 + ac - 2ac + c^2}{a+c} = \frac{a^2 - ac + c^2}{a+c} \]

4. **Combining the logarithms**:
Now we can combine the logarithms:
\[ \log(a+c) + \log\left(\frac{a^2 - ac + c^2}{a+c}\right) = \log\left((a+c) \cdot \frac{a^2 - ac + c^2}{a+c}\right) \]
The \( a+c \) cancels:
\[ = \log(a^2 - ac + c^2) \]

5. **Final Result**:
Therefore, the final expression simplifies to:
\[ \log(a^2 - ac + c^2) \]

### Final Answer:
\[ \log(a^2 - ac + c^2) \]