Question

If $a$, $b$, and $c$ are in GP, and $\log a - \log 2b$, $\log 2b - \log 3c$, $\log 3c - \log a$ are in A.P., then $a$, $b$, and $c$ are the lengths of the sides of a triangle which is

• A. acute angled
• B. obtuse angled
• C. right angled
• D. equilateral

Answer 1

The Correct Option is B

Since \( a \), \( b \), and \( c \) are in geometric progression (GP), we have the relation: \[ \frac{b}{a} = \frac{c}{b} \quad \Rightarrow \quad b^2 = ac \] This is the first key equation. Next, we are given that the three differences of logarithms: \[ \log a - \log 2b, \quad \log 2b - \log 3c, \quad \log 3c - \log a \] are in arithmetic progression. For these differences to be in A.P., the middle term must be the average of the other two. Therefore: \[ \log 2b - \log 3c = \frac{(\log a - \log 2b) + (\log 3c - \log a)}{2} \] Simplifying the equation: \[ \log 2b - \log 3c = \frac{\log a - \log 2b + \log 3c - \log a}{2} \] \[ \Rightarrow 2(\log 2b - \log 3c) = \log 3c - \log 2b \] \[ \Rightarrow 3(\log 2b - \log 3c) = 0 \] \[ \Rightarrow \log 2b = \log 3c \] Thus, we get: \[ 2b = 3c \quad \Rightarrow \quad b = \frac{3}{2}c \] We also know that \( b^2 = ac \), so substituting \( b = \frac{3}{2}c \) into this: \[ \left( \frac{3}{2}c \right)^2 = ac \] \[ \Rightarrow \frac{9}{4}c^2 = ac \] \[ \Rightarrow 9c^2 = 4ac \] \[ \Rightarrow a = \frac{9}{4}c \] So, we have the relations: \[ a = \frac{9}{4}c \quad \text{and} \quad b = \frac{3}{2}c \]

Step 2: Triangle Inequality

For \( a \), \( b \), and \( c \) to form the sides of a triangle, they must satisfy the triangle inequality: 1. \( a + b > c \) 2. \( b + c > a \) 3. \( c + a > b \) Let's check these inequalities using \( a = \frac{9}{4}c \) and \( b = \frac{3}{2}c \): 1. \( a + b > c \): \[ \frac{9}{4}c + \frac{3}{2}c > c \] \[ \Rightarrow \frac{9}{4}c + \frac{6}{4}c > c \quad \Rightarrow \quad \frac{15}{4}c > c \quad \Rightarrow \quad 15 > 4 \] This inequality is true. 2. \( b + c > a \): \[ \frac{3}{2}c + c > \frac{9}{4}c \] \[ \Rightarrow \frac{5}{2}c > \frac{9}{4}c \quad \Rightarrow \quad 10 > 9 \] This inequality is true as well. 3. \( c + a > b \): \[ c + \frac{9}{4}c > \frac{3}{2}c \] \[ \Rightarrow \frac{13}{4}c > \frac{3}{2}c \quad \Rightarrow \quad 13 > 6 \] This inequality is true.

Step 3: Type of Triangle

From the relations \( a = \frac{9}{4}c \) and \( b = \frac{3}{2}c \), the sides of the triangle satisfy the triangle inequality. Now, check if the triangle is **right-angled**. For a right-angled triangle, the Pythagorean theorem should hold: \[ a^2 + b^2 = c^2 \] Substitute \( a = \frac{9}{4}c \) and \( b = \frac{3}{2}c \) into this equation: \[ \left( \frac{9}{4}c \right)^2 + \left( \frac{3}{2}c \right)^2 = c^2 \] \[ \Rightarrow \frac{81}{16}c^2 + \frac{9}{4}c^2 = c^2 \] \[ \Rightarrow \frac{81}{16}c^2 + \frac{36}{16}c^2 = c^2 \] \[ \Rightarrow \frac{117}{16}c^2 = c^2 \] \[ \Rightarrow 117 = 16 \] This is clearly not true. Therefore, the triangle is **not right-angled**. Given that the Pythagorean theorem does not hold, but the triangle inequality does, the triangle formed by \( a \), \( b \), and \( c \) is **obtuse-angled**.

Answer:

\[ \boxed{\text{Obtuse angled}} \]