Question

The sum of the squares of three distinct real numbers, which are in G.P. is $S^2$. If their sum is $\alpha S$ and if ${\alpha }^2ϵ(a,b)-\left\{c\right\}$, then find the value of ab + c. ___

Answer 1

Let the numbers be $ar,a,\frac{a}{r}$ such that $a(r+1+\frac{1}{r})=\alpha S$

and $a^2(r^2+1+\frac{1}{r^2})=S^2$

Put $r+\frac{1}{r}=t\therefore r^2+\frac{1}{r^2}=t^2-2$

$\therefore a(t+1)=\alpha S$ and

$a^2(t^2-1)=S^2$

Eliminating S, we get $a^2(t^2-1)=\frac{a^2(t+1{)}^2}{{\alpha }^2}$

$\therefore (t-1){\alpha }^2=(t+1)$

or $t=\frac{{\alpha }^2+1}{{\alpha }^2-1}$

Now $t=r+\frac{1}{r}\therefore r^2-rt+1=0$

For t to be real $t^2-4>0\therefore (t+2)(t-2)>0$

$\therefore t<-2ort>2$

Hence from (1), we get $\frac{{\alpha }^2+1}{{\alpha }^2-1}<-2$

or $\frac{{\alpha }^2+1}{{\alpha }^2-1}>2$

$\Rightarrow \frac{{\alpha }^2+1}{{\alpha }^2-1}+2<0$

or $\frac{{\alpha }^2+1}{{\alpha }^2-1}-2>0$

$\frac{{\alpha }^2-\frac{1}{3}}{{\alpha }^2-1}<0$

or $\frac{{\alpha }^2-3}{{\alpha }^2-1}<0$

$\therefore \frac{1}{3}<{\alpha }^2<1$

or $1<{\alpha }^2<3$

$\therefore {\alpha }^2ϵ(\frac{1}{3},1)\cup (1,3)\Rightarrow {\alpha }^2ϵ(\frac{1}{3},3)-\left\{1\right\}$

$\therefore \alpha =\frac{1}{3},b=3,c=1$

$ab+c=2$