Key Concepts and Formulas

1. Arithmetic Progression (A.P.): For three numbers $x, y, z$ in A.P., $2y = x + z$.
2. Geometric Progression (G.P.): For three numbers $x, y, z$ in G.P., $y^2 = xz$.
3. Quadratic Equation: The roots of $Ax^2 + Bx + C = 0$ are $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. If $x_1, x_2$ are roots, the equation is $x^2 - (x_1+x_2)x + x_1x_2 = 0$.

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Step-by-Step Solution

Step 1: Utilize the A.P. property and the sum to find the value of 'b'. We are given that $a, b, c$ are in A.P. This means the middle term $b$ is the arithmetic mean of $a$ and $c$, so $2b = a + c$. We are also given the sum $a + b + c = \frac{3}{4}$. Substitute $a + c = 2b$ into the sum equation: $(a+c) + b = \frac{3}{4}$ $2b + b = \frac{3}{4}$ $3b = \frac{3}{4}$ Dividing by 3, we get: $b = \frac{1}{4}$

Step 2: Utilize the G.P. property to establish a relationship between 'a' and 'c'. We are given that $a^2, b^2, c^2$ are in G.P. This means the square of the middle term is equal to the product of the other two terms: $(b^2)^2 = a^2 c^2$ $b^4 = (ac)^2$ Taking the square root of both sides, we get: $b^2 = \pm ac$ Substitute the value of $b = \frac{1}{4}$: $(\frac{1}{4})^2 = \pm ac$ $\frac{1}{16} = \pm ac$ This gives us two possibilities for the product $ac$: Case 1: $ac = \frac{1}{16}$ Case 2: $ac = -\frac{1}{16}$

Step 3: Formulate quadratic equations for 'a' and 'c'. From Step 1, we know $a + c = 2b = 2(\frac{1}{4}) = \frac{1}{2}$. We now have the sum ($a+c = \frac{1}{2}$) and two possible products ($ac = \frac{1}{16}$ or $ac = -\frac{1}{16}$) for $a$ and $c$. The numbers $a$ and $c$ are the roots of a quadratic equation of the form $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.

Case 1: $ac = \frac{1}{16}$ The quadratic equation is: $x^2 - \frac{1}{2}x + \frac{1}{16} = 0$ Multiplying by 16 to clear fractions: $16x^2 - 8x + 1 = 0$ This is a perfect square trinomial: $(4x - 1)^2 = 0$ $4x - 1 = 0 \implies x = \frac{1}{4}$ In this case, both roots are $\frac{1}{4}$. So, $a = \frac{1}{4}$ and $c = \frac{1}{4}$. This contradicts the condition $a < b < c$ since $a=b=c$. Therefore, this case is invalid.

Case 2: $ac = -\frac{1}{16}$ The quadratic equation is: $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$ Multiplying by 16 to clear fractions: $16x^2 - 8x - 1 = 0$ Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ with $A=16, B=-8, C=-1$: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(16)(-1)}}{2(16)}$ $x = \frac{8 \pm \sqrt{64 + 64}}{32}$ $x = \frac{8 \pm \sqrt{128}}{32}$ $x = \frac{8 \pm 8\sqrt{2}}{32}$ $x = \frac{1 \pm \sqrt{2}}{4}$ The two possible values for $a$ (or $c$) are $\frac{1 + \sqrt{2}}{4}$ and $\frac{1 - \sqrt{2}}{4}$.

Step 4: Apply the condition $a < b < c$ to find the correct value of 'a'. We have $b = \frac{1}{4}$. The two potential values for $a$ are $\frac{1 + \sqrt{2}}{4}$ and $\frac{1 - \sqrt{2}}{4}$.

Option 1: $a = \frac{1 + \sqrt{2}}{4}$ Since $\sqrt{2} \approx 1.414$, $a \approx \frac{1 + 1.414}{4} = \frac{2.414}{4} \approx 0.6035$. Our value of $b$ is $\frac{1}{4} = 0.25$. Here, $a > b$ ($0.6035 > 0.25$), which violates the condition $a < b < c$. So, this option is incorrect.

Option 2: $a = \frac{1 - \sqrt{2}}{4}$ Since $\sqrt{2} \approx 1.414$, $a \approx \frac{1 - 1.414}{4} = \frac{-0.414}{4} \approx -0.1035$. Our value of $b$ is $\frac{1}{4} = 0.25$. Here, $a < b$ ($-0.1035 < 0.25$), which satisfies the first part of the condition. Now, let's find $c$. We know $a + c = \frac{1}{2}$, so $c = \frac{1}{2} - a$. $c = \frac{1}{2} - \left(\frac{1 - \sqrt{2}}{4}\right)$ $c = \frac{2}{4} - \frac{1 - \sqrt{2}}{4}$ $c = \frac{2 - (1 - \sqrt{2})}{4}$ $c = \frac{1 + \sqrt{2}}{4}$ We have $a = \frac{1 - \sqrt{2}}{4}$, $b = \frac{1}{4}$, and $c = \frac{1 + \sqrt{2}}{4}$. Checking the condition $a < b < c$: $a \approx -0.1035$ $b = 0.25$ $c \approx 0.6035$ This clearly satisfies $a < b < c$.

Therefore, the value of $a$ is $\frac{1 - \sqrt{2}}{4}$.

Common Mistakes & Tips

• Sign errors when taking square roots: Remember that $\sqrt{x^2} = |x|$. In this problem, $b^2 = \pm ac$ leads to two distinct cases.
• Forgetting the $a < b < c$ condition: This condition is crucial for eliminating extraneous solutions. Always verify the final answer against all given constraints.
• Algebraic manipulation: Be careful with fractions and signs when solving quadratic equations.

Summary We used the properties of A.P. and G.P. to establish relationships between $a, b,$ and $c$. The A.P. property and the given sum allowed us to find $b$. The G.P. property led to two possible values for the product $ac$. By forming quadratic equations with the sum and product of $a$ and $c$, we found potential values for $a$. Finally, we applied the condition $a < b < c$ to select the correct value of $a$.

The value of $a$ is $\frac{1 - \sqrt{2}}{4}$, which matches option (A).

The final answer is $\boxed{\text{A}}$.