Arithmetic Progression (AP) Conditions

Given that a, b, c are in Arithmetic Progression (A.P.), the relationship between them is:

$2b = a + c$

Sum Condition

We are also given that the sum of the terms is 1:

$a + b + c = 1$

Substitute the A.P. condition ($a + c = 2b$) into the sum equation:

$ (a + c) + b = 1 $

$ 2b + b = 1 $

$ 3b = 1 $

$ b = \frac{1}{3} $

Geometric Progression (GP) Conditions

The terms $a^2, 2b^2, c^2$ are in Geometric Progression (G.P.). This implies:

$ (2b^2)^2 = a^2 \times c^2 $

$ 4b^4 = (ac)^2 $

Taking the square root of both sides gives two possibilities:

$ 2b^2 = ac \quad \text{or} \quad 2b^2 = -ac $

Evaluating GP Possibilities

Substitute $b = \frac{1}{3}$ into these equations:

$ 2\left(\frac{1}{3}\right)^2 = 2\left(\frac{1}{9}\right) = \frac{2}{9} $

Case 1: $ac = \frac{2}{9}$. We also know $a+c = 2b = \frac{2}{3}$. The quadratic equation $x^2 - (a+c)x + ac = 0$ becomes $x^2 - \frac{2}{3}x + \frac{2}{9} = 0$, or $9x^2 - 6x + 2 = 0$. The discriminant is $D = (-6)^2 - 4(9)(2) = 36 - 72 = -36$. This yields complex roots, which contradicts the condition $a < b < c$ (implying real numbers).

Case 2: $-ac = \frac{2}{9}$, which means $ac = -\frac{2}{9}$. We still have $a+c = \frac{2}{3}$. The quadratic equation $x^2 - (a+c)x + ac = 0$ becomes $x^2 - \frac{2}{3}x - \frac{2}{9} = 0$, or $9x^2 - 6x - 2 = 0$.

Finding a and c

Solve the quadratic equation $9x^2 - 6x - 2 = 0$ using the quadratic formula:

$ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(9)(-2)}}{2(9)} = \frac{6 \pm \sqrt{36 + 72}}{18} = \frac{6 \pm \sqrt{108}}{18} $

$ x = \frac{6 \pm 6\sqrt{3}}{18} = \frac{1 \pm \sqrt{3}}{3} $

The values for a and c are $\frac{1 - \sqrt{3}}{3}$ and $\frac{1 + \sqrt{3}}{3}$.

Given $a < b < c$, we have:

$ a = \frac{1 - \sqrt{3}}{3}, \quad b = \frac{1}{3}, \quad c = \frac{1 + \sqrt{3}}{3} $

This satisfies $a < b < c$ since $\sqrt{3} \approx 1.732$, making $a$ negative and $c$ positive.

Calculating $a^2 + b^2 + c^2$

Calculate the squares of a, b, and c:

• $ a^2 = \left(\frac{1 - \sqrt{3}}{3}\right)^2 = \frac{1 - 2\sqrt{3} + 3}{9} = \frac{4 - 2\sqrt{3}}{9} $
• $ b^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} $
• $ c^2 = \left(\frac{1 + \sqrt{3}}{3}\right)^2 = \frac{1 + 2\sqrt{3} + 3}{9} = \frac{4 + 2\sqrt{3}}{9} $

Sum the squares:

$ a^2 + b^2 + c^2 = \left(\frac{4 - 2\sqrt{3}}{9}\right) + \left(\frac{1}{9}\right) + \left(\frac{4 + 2\sqrt{3}}{9}\right) $

$ a^2 + b^2 + c^2 = \frac{4 - 2\sqrt{3} + 1 + 4 + 2\sqrt{3}}{9} = \frac{9}{9} = 1 $

Final Calculation

The question asks for the value of $9(a^2 + b^2 + c^2)$.

$ 9(a^2 + b^2 + c^2) = 9(1) = 9 $