ISI2021-MMA: 11

Question

Suppose $a, b, c$ are in $\text{A.P.}$ and $a^{2}, b^{2}, c^{2}$ are in $\text{G.P.}$ 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{2}}$
• C. $\frac{1}{2} – \frac{1}{\sqrt{3}}$
• D. $\frac{1}{2} – \frac{1}{\sqrt{2}}$

Answer 1

Answer: (D)

$a, b, c$ are in AP. Let $a = x-d$, $b = x$, $c = x+d$ where $d$ is the common difference.

Given, $a+b+c = \frac{3}{2}$

            $x – d + x + x + d = \frac{3}{2} \implies 3x = \frac{3}{2} \implies x = \frac{1}{2}$

$a^2, b^2, c^2$ are in GP.

$\therefore \frac{b^2}{a^2} = \frac{c^2}{b^2} \implies b^4 = (ac)^2$

                                 $x^4 = ((x-d)(x+d))^2 \implies x^4 = (x^2 – d^2)^2$

                                 $x^4 = x^4 + d^4 – 2x^2d^2 \implies d^4 – 2x^2d^2 = 0$

                                 $d^2(d^2 – 2x^2) = 0$

                                Since $a < b < c, \therefore d\neq0$

                                Hence, $d^2 – 2x^2 = 0 \implies d^2 = 2*\frac{1}{4} \implies d = \frac{1}{\sqrt{2}}$

Now we know $a = x – d = \frac{1}{2} – \frac{1}{\sqrt{2}}$