Question

If \( a, b, c \) are in A.P. where \( a + b + c = 1 \) and \( a, 2b, c \) are in G.P., then the value of \( 9(a^2 + b^2 + c^2) \) is equal to:

• A. 3
• B. -3
• C. 4
• D. -4

Hint:

For problems involving A.P. and G.P., use the known properties of sequences and solve the system of equations step by step.

Answer 1

The Correct Option is B

Given that \(a, b, c\) are in Arithmetic Progression (A.P.), \(a, 2b, c\) are in Geometric Progression (G.P.), and \(a + b + c = 1\).

1. Since \(a, b, c\) are in A.P.:

   \(2b = a + c\)

2. Since \(a, 2b, c\) are in G.P.:

   \((2b)^2 = ac \Rightarrow 4b^2 = ac\)

3. Using the given condition:

   \(a + b + c = 1\)

   Substituting \(a + c = 2b\):

   \(2b + b = 1 \Rightarrow 3b = 1 \Rightarrow b = \dfrac{1}{3}\)

4. Substitute \(b = \dfrac{1}{3}\) into previous relations:

   \(a + c = \dfrac{2}{3}\)

   \(ac = 4b^2 = 4\left(\dfrac{1}{3}\right)^2 = \dfrac{4}{9}\)

5. Now evaluate:

   \(9(a^2 + b^2 + c^2)\)

   Using the identity \(a^2 + c^2 = (a + c)^2 - 2ac\):

   \(a^2 + c^2 = \left(\dfrac{2}{3}\right)^2 - 2\left(\dfrac{4}{9}\right) = \dfrac{4}{9} - \dfrac{8}{9} = -\dfrac{4}{9}\)

6. Add \(b^2\):

   \(a^2 + b^2 + c^2 = -\dfrac{4}{9} + \dfrac{1}{9} = -\dfrac{3}{9} = -\dfrac{1}{3}\)

7. Multiply by 9:

   \(9(a^2 + b^2 + c^2) = 9 \times \left(-\dfrac{1}{3}\right) = -3\)

Therefore, the correct answer is:

\(\boxed{-3}\)