Question

Let $a,b,c$ be in A.P. and let $a^2,b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=\frac{3}{2}$, then $a=$

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

Answer 1

The correct option is D $\frac{1}{2}-$ $\frac{1}{\sqrt{2}}$

If $a,b,c$ are in AP then,

$2b=a+c$

$b=\frac{1}{2}(a+c)$ … (i)

If $a^2,b^2,c^2$ are in GP then $b^4=a^2{c}^2$ or $b^2=\pm ac$ … (ii)

Using (i) in $a+b+c=\frac{3}{2}$ gives $3b=\frac{3}{2}$

$b=\frac{1}{2}$

Hence $a+c=1$ & $ac=\pm \frac{1}{4}$

So $a$ & $c$ are roots of either $x^2-x+\frac{1}{4}=0$ or $x^2-x-\frac{1}{4}=0$

The first has equal roots of $x=\frac{1}{2}$ and second gives $x==\frac{1}{2}1\pm \sqrt{2}$ for $a$ & $c$

Since $a<c$ we must have $x=\frac{1}{2}(1-\sqrt{2})$