Question

Suppose a,b,c are in A.P. and $a^2, b^2, c^2$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$. Then the value of a is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{2} - \frac{1}{\sqrt{2}}$
• D. $\frac{1}{2} - \frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{1}{2} - \frac{1}{\sqrt{2}}$

Answer 2

Given $a, b, c$ in AP, we have $2b = a+c$. Given $a+b+c = \frac{3}{2}$. Substituting $a+c=2b$, we get $2b+b = \frac{3}{2}$, which simplifies to $3b = \frac{3}{2}$, so $b = \frac{1}{2}$. From $a+c=2b$, we have $a+c = 2(\frac{1}{2}) = 1$.

Given $a^2, b^2, c^2$ in GP, we have $(b^2)^2 = a^2 c^2$, which means $b^4 = a^2 c^2$. Taking the square root of both sides, $b^2 = \pm ac$. Since $b = \frac{1}{2}$, $b^2 = \frac{1}{4}$. So, $\frac{1}{4} = \pm ac$.

Case 1: $\frac{1}{4} = ac$. We have the system of equations: $a+c = 1$ $ac = \frac{1}{4}$ The quadratic equation with roots $a$ and $c$ is $x^2 - (a+c)x + ac = 0$, which becomes $x^2 - x + \frac{1}{4} = 0$. Multiplying by 4, we get $4x^2 - 4x + 1 = 0$, or $(2x-1)^2 = 0$. This gives $x = \frac{1}{2}$. So, $a = c = \frac{1}{2}$. This contradicts the condition $a < b < c$.

Case 2: $\frac{1}{4} = -ac$. So, $ac = -\frac{1}{4}$. We have the system of equations: $a+c = 1$ $ac = -\frac{1}{4}$ The quadratic equation with roots $a$ and $c$ is $x^2 - (a+c)x + ac = 0$, which becomes $x^2 - x - \frac{1}{4} = 0$. Multiplying by 4, we get $4x^2 - 4x - 1 = 0$. Using the quadratic formula, $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)} = \frac{4 \pm \sqrt{16+16}}{8} = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$. The two roots are $\frac{1 + \sqrt{2}}{2}$ and $\frac{1 - \sqrt{2}}{2}$. Given $a < b < c$ and $b = \frac{1}{2}$, we must have $a < \frac{1}{2}$ and $c > \frac{1}{2}$. Therefore, $a = \frac{1 - \sqrt{2}}{2}$ and $c = \frac{1 + \sqrt{2}}{2}$. The value of $a$ is $\frac{1 - \sqrt{2}}{2} = \frac{1}{2} - \frac{\sqrt{2}}{2} = \frac{1}{2} - \frac{1}{\sqrt{2}}$.