If `log(a+c)+log(a+c-2b)=2log(a-c)` then

Options

A. a,b,c are in A.P.
B. a,b,c are in G.P.
C. a,b,c are in H.P.
D. a+b, b+c , c+a are in A.P.

Answer

To solve the equation \( \log(a+c) + \log(a+c-2b) = 2\log(a-c) \), we will use properties of logarithms step by step.

### Step 1: Use the property of logarithms
We can combine the left-hand side using the property \( \log A + \log B = \log(AB) \).

\[ \log((a+c)(a+c-2b)) = 2\log(a-c) \]

### Step 2: Rewrite the right-hand side
Using the property \( n\log A = \log(A^n) \), we can rewrite the right-hand side:

\[ \log((a+c)(a+c-2b)) = \log((a-c)^2) \]

### Step 3: Eliminate the logarithm
Since the logarithms are equal, we can set the arguments equal to each other:

\[ (a+c)(a+c-2b) = (a-c)^2 \]

### Step 4: Expand both sides
Now, let's expand both sides of the equation.

Left-hand side:
\[ (a+c)(a+c-2b) = a^2 + ac - 2ab + ac + c^2 - 2bc = a^2 + 2ac + c^2 - 2ab - 2bc \]

Right-hand side:
\[ (a-c)^2 = a^2 - 2ac + c^2 \]

### Step 5: Set the expanded forms equal
Now we have:

\[ a^2 + 2ac + c^2 - 2ab - 2bc = a^2 - 2ac + c^2 \]

### Step 6: Simplify the equation
Subtract \( a^2 + c^2 \) from both sides:

\[ 2ac - 2ab - 2bc = -2ac \]

### Step 7: Combine like terms
Adding \( 2ac \) to both sides gives:

\[ 4ac - 2ab - 2bc = 0 \]

### Step 8: Factor out common terms
We can factor out 2 from the left side:

\[ 2(2ac - ab - bc) = 0 \]

### Step 9: Set the factor equal to zero
This gives us:

\[ 2ac - ab - bc = 0 \]

### Step 10: Rearranging gives the final result
Rearranging this leads to:

\[ ab + bc = 2ac \]

This is the condition for \( a, b, c \) to be in Harmonic Progression (HP).

### Final Result
Thus, we conclude that \( a, b, c \) are in Harmonic Progression (HP).

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