If `4a^(2)+9b^(2)+16c^(2)=2(3ab+6bc+4ca)," where "a,b,c` are non-zero numbers, then a,b,c are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the equation \( 4a^2 + 9b^2 + 16c^2 = 2(3ab + 6bc + 4ca) \) and determine the relationship between \( a, b, c \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 4a^2 + 9b^2 + 16c^2 = 2(3ab + 6bc + 4ca) \]
Expanding the right-hand side:
\[ 4a^2 + 9b^2 + 16c^2 = 6ab + 12bc + 8ca \]

### Step 2: Move all terms to one side
Rearranging the equation gives us:
\[ 4a^2 + 9b^2 + 16c^2 - 6ab - 12bc - 8ca = 0 \]

### Step 3: Group terms for completing the square
We can group the terms to complete the square:
\[ (4a^2 - 6ab + 9b^2) + (16c^2 - 12bc - 8ca) = 0 \]

### Step 4: Complete the square for each group
1. For \( 4a^2 - 6ab + 9b^2 \):
\[ 4a^2 - 6ab + 9b^2 = (2a - 3b)^2 \]

2. For \( 16c^2 - 12bc - 8ca \):
\[ 16c^2 - 12bc - 8ca = (4c - 3b)^2 + (4c - 2a)^2 \]

Thus, the equation becomes:
\[ (2a - 3b)^2 + (3b - 4c)^2 + (4c - 2a)^2 = 0 \]

### Step 5: Analyze the equation
The sum of squares equals zero only when each individual square is zero:
1. \( 2a - 3b = 0 \) (i.e., \( 2a = 3b \))
2. \( 3b - 4c = 0 \) (i.e., \( 3b = 4c \))
3. \( 4c - 2a = 0 \) (i.e., \( 4c = 2a \))

### Step 6: Solve for relationships
From \( 2a = 3b \), we can express \( a \) in terms of \( b \):
\[ a = \frac{3}{2}b \]

From \( 3b = 4c \), we can express \( b \) in terms of \( c \):
\[ b = \frac{4}{3}c \]

Substituting \( b \) into the expression for \( a \):
\[ a = \frac{3}{2} \left(\frac{4}{3}c\right) = 2c \]

### Step 7: Express \( a, b, c \) in terms of a common variable
Let \( c = x \):
- Then \( a = 2x \)
- And \( b = \frac{4}{3}x \)

### Step 8: Check the relationship
We have \( a, b, c \) as:
- \( a = 2x \)
- \( b = \frac{4}{3}x \)
- \( c = x \)

### Step 9: Determine the type of progression
The ratios of \( a, b, c \) are:
- \( \frac{a}{b} = \frac{2x}{\frac{4}{3}x} = \frac{3}{2} \)
- \( \frac{b}{c} = \frac{\frac{4}{3}x}{x} = \frac{4}{3} \)

Since \( 2, \frac{4}{3}, 3 \) are in arithmetic progression, we can conclude that \( a, b, c \) are in harmonic progression (HP).

### Final Conclusion
Thus, the numbers \( a, b, c \) are in HP.

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