Question

If $a, b, c$ are real numbers such that $(a-2)^{2}+(b-2)^{2}+(c-2)^{2}=0$, then ________.

• A. $a, b, c$ are in G.P. and $a+b+c=6$
• B. $a, b, c$ are in G.P. and $a+b+c=4$
• C. $a, b, c$ are not in G.P.
• D. $a, b, c$ are in G.P. and $a+b+c=8$
• E. $a, b, c$ are not in G.P. and $a+b+c=16$

Hint:

$x^2+y^2+z^2=0 \implies x=y=z=0$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The problem is based on a fundamental property of real numbers: the square of any real number is non-negative (i.e., greater than or equal to zero). If a sum of squares of real numbers equals zero, then each individual term must be zero.
Step 2: Key Formula or Approach
If \(x, y, z\) are real numbers, and \(x^2 + y^2 + z^2 = 0\), then it must be that \(x=0\), \(y=0\), and \(z=0\).
Step 3: Detailed Explanation
1. Apply the property of the sum of squares.
We are given the equation \((a - 2)^2 + (b - 2)^2 + (c - 2)^2 = 0\).
Since \(a, b, c\) are real numbers, \((a-2)\), \((b-2)\), and \((c-2)\) are also real.
Their squares, \((a-2)^2\), \((b-2)^2\), and \((c-2)^2\), are all non-negative.
The only way for their sum to be zero is if each term is individually zero.
\[ (a - 2)^2 = 0 \implies a - 2 = 0 \implies a = 2 \] \[ (b - 2)^2 = 0 \implies b - 2 = 0 \implies b = 2 \] \[ (c - 2)^2 = 0 \implies c - 2 = 0 \implies c = 2 \] So, we have found that \(a = b = c = 2\).
2. Check the conditions in the options.
Now we must check which of the given options is consistent with these values.
- Check if a, b, c are in Geometric Progression (G.P.):
A sequence is in G.P. if the ratio between consecutive terms is constant (the common ratio, r).
Common ratio \(r = \frac{b}{a} = \frac{2}{2} = 1\).
Also, \(\frac{c}{b} = \frac{2}{2} = 1\).
Since the ratio is constant, the numbers 2, 2, 2 are in a G.P. with a common ratio of 1.
- Check the sum a + b + c:
\(a + b + c = 2 + 2 + 2 = 6\).
3. Evaluate the options.
- (A) a, b, c are in G.P. and a + b + c = 6. This is TRUE.
- (B) a, b, c are in G.P. and a + b + c = 4. This is FALSE (the sum is 6).
- (C) a, b, c are not in G.P. This is FALSE.
- (D) a, b, c are in G.P. and a + b + c = 8. This is FALSE (the sum is 6).
- (E) a, b, c are not in G.P. and a + b + c = 16. This is FALSE.
Step 4: Final Answer
The only correct statement is that a, b, c are in G.P. and their sum is 6.