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 $1, 3+2\sqrt{2} , 3- 2\sqrt{2}$.