If `21(a^(2) + b^(2) + c^(2)) = (a + 2b + 4c)^(2)` 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 \( 21(a^2 + b^2 + c^2) = (a + 2b + 4c)^2 \) and determine the relationship between \( a, b, c \), we can follow these steps:

### Step 1: Expand the Right-Hand Side
First, we need to expand the right-hand side of the equation:

\[ (a + 2b + 4c)^2 = a^2 + (2b)^2 + (4c)^2 + 2(a)(2b) + 2(a)(4c) + 2(2b)(4c) \]

This simplifies to:

\[ = a^2 + 4b^2 + 16c^2 + 4ab + 8ac + 16bc \]

### Step 2: Set Up the Equation
Now, we can rewrite the original equation with the expanded form:

\[ 21(a^2 + b^2 + c^2) = a^2 + 4b^2 + 16c^2 + 4ab + 8ac + 16bc \]

### Step 3: Rearranging the Equation
Rearranging gives us:

\[ 21a^2 + 21b^2 + 21c^2 - (a^2 + 4b^2 + 16c^2 + 4ab + 8ac + 16bc) = 0 \]

Combining like terms:

\[ (21a^2 - a^2) + (21b^2 - 4b^2) + (21c^2 - 16c^2) - 4ab - 8ac - 16bc = 0 \]

This simplifies to:

\[ 20a^2 + 17b^2 + 5c^2 - 4ab - 8ac - 16bc = 0 \]

### Step 4: Applying Lagrange's Identity
We can apply Lagrange's identity here. According to the identity:

\[ L_1^2 + M_1^2 + N_1^2 = L_2^2 + M_2^2 + N_2^2 \]

We can identify:

- \( L_1 = 1, M_1 = 2, N_1 = 4 \)
- \( L_2 = a, M_2 = b, N_2 = c \)

This leads to:

\[ 21(1^2 + 2^2 + 4^2)(a^2 + b^2 + c^2) - (a + 2b + 4c)^2 = 0 \]

### Step 5: Setting Up Conditions
From the equation \( 20a^2 + 17b^2 + 5c^2 - 4ab - 8ac - 16bc = 0 \), we can derive the following conditions:

1. \( b - 2a = 0 \) (i.e., \( b = 2a \))
2. \( c - 2b = 0 \) (i.e., \( c = 2b \))
3. \( c - 4a = 0 \) (i.e., \( c = 4a \))

### Step 6: Expressing in Terms of a Single Variable
From these relationships, we can express \( a, b, c \) in terms of a single variable \( k \):

- Let \( a = k \)
- Then \( b = 2k \)
- And \( c = 4k \)

### Conclusion
Thus, we have established that \( a, b, c \) are in a geometric progression (GP) with a common ratio of 2.

### Final Answer
Therefore, \( a, b, c \) are in GP.

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