`a, b, c` are sides of a triangle and `a, b, c` are in GP If `log a- log 2b, log 2b-log 3c and log 3c-log a` are in AP then

Options

A. acute angled
B. obtuse angled
C. right angled
D. none of these

Answer

To solve the problem step by step, we will follow the logical deductions as outlined in the video transcript.

### Step 1: Understand the Given Conditions
We are given that \( a, b, c \) are the sides of a triangle and they are in geometric progression (GP). This means that:
\[ b^2 = ac \tag{1} \]

### Step 2: Set Up the Arithmetic Progression (AP) Condition
We are also given that:
\[ \log a - \log 2b, \quad \log 2b - \log 3c, \quad \log 3c - \log a \]
are in arithmetic progression (AP). For three numbers \( x, y, z \) to be in AP, the condition is:
\[ 2y = x + z \]
Applying this to our logs, we have:
\[ 2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a) \]

### Step 3: Simplify the AP Condition
Substituting the logs into the equation:
\[ 2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a) \]
This simplifies to:
\[ 2\log \frac{2b}{3c} = \log \frac{a}{2b} + \log \frac{3c}{a} \]
Using the property of logarithms \( \log x + \log y = \log (xy) \):
\[ 2\log \frac{2b}{3c} = \log \left(\frac{a \cdot 3c}{2b \cdot a}\right) = \log \frac{3c}{2b} \]

### Step 4: Exponentiate to Remove Logarithms
Exponentiating both sides gives:
\[ \left(\frac{2b}{3c}\right)^2 = \frac{3c}{2b} \]
Cross-multiplying yields:
\[ 4b^2 = 9c^2 \tag{2} \]

### Step 5: Relate \( b \) and \( c \)
From equation (2), we can express \( b \) in terms of \( c \):
\[ \frac{b^2}{c^2} = \frac{9}{4} \implies b = \frac{3}{2}c \]

### Step 6: Substitute \( b \) into Equation (1)
Now substituting \( b = \frac{3}{2}c \) into equation (1):
\[ \left(\frac{3}{2}c\right)^2 = ac \implies \frac{9}{4}c^2 = ac \]
This simplifies to:
\[ a = \frac{9}{4}c \]

### Step 7: Determine the Relationship Between \( a, b, c \)
Now we have:
- \( a = \frac{9}{4}c \)
- \( b = \frac{3}{2}c \)

### Step 8: Check the Triangle Inequality
For \( a, b, c \) to form a triangle, they must satisfy the triangle inequalities:
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

Substituting the values:
1. \( \frac{9}{4}c + \frac{3}{2}c > c \) simplifies to \( \frac{9}{4}c + \frac{6}{4}c > \frac{4}{4}c \) which is true.
2. \( \frac{9}{4}c + c > \frac{3}{2}c \) simplifies to \( \frac{9}{4}c + \frac{4}{4}c > \frac{6}{4}c \) which is true.
3. \( \frac{3}{2}c + c > \frac{9}{4}c \) simplifies to \( \frac{6}{4}c + \frac{4}{4}c > \frac{9}{4}c \) which is true.

### Step 9: Determine the Type of Triangle
Now we can find \( b^2 + c^2 \) and compare it with \( a^2 \):
\[ b^2 + c^2 = \left(\frac{3}{2}c\right)^2 + c^2 = \frac{9}{4}c^2 + c^2 = \frac{9}{4}c^2 + \frac{4}{4}c^2 = \frac{13}{4}c^2 \]
\[ a^2 = \left(\frac{9}{4}c\right)^2 = \frac{81}{16}c^2 \]
To compare:
\[ \frac{13}{4}c^2 = \frac{52}{16}c^2 \]
Since \( \frac{81}{16}c^2 > \frac{52}{16}c^2 \), we have:
\[ a^2 > b^2 + c^2 \]
This indicates that the triangle is obtuse.

### Conclusion
Thus, the angle opposite to side \( a \) is obtuse.