Question

Three non-zero real numbers form an $A.P$. and the square of the numbers taken in the same order constitute a $G.P$. Then the number of all possible common ratios of the $G.P$. are

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

Three numbers in $A.P$. can be taken as $a-d,a,a+d$
Then ${\left(a-d\right)}^2,a^2,{\left(a+d\right)}^2$ are in $G.P$.
$\Rightarrow a^4={\left(a^2-d^2\right)}^2$
$\Rightarrow d^4-2a^2{d}^2=0$
$\Rightarrow d^2\left(d^2-2a^2\right)=0$
$\Rightarrow d=0,\pm \sqrt{2a}$
$\because {\left(a-d\right)}^2,a^2,{\left(a+d\right)}^2$ forms a $G.P$.
$\therefore$ Common ratio $\left(r\right)={\left(\frac{a+d}{a}\right)}^2$
When $d=0$,
$r={\left(\frac{a+d}{a}\right)}^2=1$
When $d=\pm \sqrt{2a},r$
$={\left(\frac{a\pm \sqrt{2a}}{a}\right)}^2$
$={\left(1\pm \sqrt{2}\right)}^2$
$=3\pm 2\sqrt{2}$
Thus, there are three common ratios $1,3+2\sqrt{2},3-2\sqrt{2}$.