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

The Correct Option is 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 $1, 3+2\sqrt{2} , 3- 2\sqrt{2}$.