Question

The sum of the squares of three distinct real numbers which are in G.P. is $S^2$. If their sum is $aS$, then ____________

• A. $1<a^2<3$
• B. $\frac{1}{3}<a^2<1$
• C. $1<a<3$
• D. $\frac{1}{3}<a<1$

Answer 1

Answer: (A), (B)

The correct options are
A $1<a^2<3$
B $\frac{1}{3}<a^2<1$

Let the three terms in GP be $x,xr,$ $x{r}^2$
$x+xr+x{r}^2=S$
$x^2+x^2{r}^2+x^2{r}^4=aS$
Therefore, $\frac{{(1+r+r^2)}^2}{1+r^2+r^4}=a^2$
$\frac{{(1+r+r^2)}^2}{(1+r+r^2)\times (1-r+r^2)}=a^2$
$\frac{(1+r+r^2)}{(1-r+r^2)}=a^2$
$(1-a^2)r^2+(1+a^2)r+(1-a^2)=0$
Since $r$ is real, ${(1+a^2)}^2-4{(1-a^2)}^2\ge 0$
Let $a^2=t$
$(1+t{)}^2-4(1-t{)}^2\ge 0$
$(3t-1)(t-3)\le 0$
From the equation, we get

$\frac{1}{3}\ge a^2\le 3$
But for $a^2=1$, $r=0$ and GP will not be defined.
Hence, $\frac{1}{3}\ge a^2<1$ and $1>a^2\le 3$