Question

Three non-zero real numbers form A.P. and the squares of these numbers taken in the same order form a G.P. Then the number of all possible common ratios of the G.P. is

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

Let the three numbers in arithmetic progression be

$a-d,a,a+d$

The squares of these numbers respectively are

${\left(a-d\right)}^2,a^2,{\left(a+d\right)}^2$

According to given condition the squares of the numbers form a geometric progression

$\Rightarrow$ ${\left(a-d\right)}^2,a^2,{\left(a+d\right)}^2$ are in a geometric progression

$\Rightarrow$ ${\left(a^2\right)}^2={\left(a-d\right)}^2{\left(a+d\right)}^2$ $\left[\because y^2=xz,whenx,y,zareinG.P\right]$

$\Rightarrow$ $a^4={\left(a^2-d^2\right)}^2$ $\left[\because \left(x+y\right)\left(x-y\right)=x^2-y^2\right]$

$\Rightarrow$ $a^2=a^2-d^2$ or $a^2=d^2-a^2$

$\Rightarrow$ $d^2=0$ or $d^2=2a^2$

$\Rightarrow$ $d=0$ or $d=\pm \sqrt{2}a$

As the given terms are distinct $d$ cannot be zero $\Rightarrow$$d=\pm \sqrt{2}a$

The common ratio $r$ is the ratio of the consecutive terms

$\Rightarrow$ $r=\frac{a^2}{{\left(a-d\right)}^2}$

$\Rightarrow$ $r=\frac{a^2}{{\left(a\pm \sqrt{2}a\right)}^2}$

$\Rightarrow$ $r=\frac{a^2}{{\left(a+\sqrt{2}a\right)}^2}$ or $r=\frac{a^2}{{\left(a-\sqrt{2}a\right)}^2}$

$\Rightarrow$ $r=\frac{1}{{\left(1+\sqrt{2}\right)}^2}$ or $r=\frac{1}{{\left(1-\sqrt{2}\right)}^2}$

Hence there are $2$ possible values of the common ratio for the squares of the numbers to be in a geometric progression.

Hence, option B is the correct answer.