Question

If three non-zero numbers are in A.P. and the squares of those numbers taken in same order are in G.P., then the common ratio can be

• A. $0$
• B. $1$
• C. $3+2\sqrt{2}$
• D. $3-2\sqrt{2}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $1$
C $3+2\sqrt{2}$
D $3-2\sqrt{2}$

Let the numbers be $a-d,a,a+d$
Now, $(a-d{)}^2,a^2,(a+d{)}^2$ are in G.P., so
$a^4=(a-d{)}^2\times (a+d{)}^2(\because B^2=AC)\Rightarrow a^4=(a^2-d^2{)}^2\Rightarrow d^4-2a^2{d}^2=0\Rightarrow d^2(d^2-2a^2)=0\Rightarrow d^2=0,2a^2\Rightarrow d=0,\pm \sqrt{2}a$

When $d=0,r=1$

When $d=\sqrt{2}a,$
$r=\frac{a^2}{(a-\sqrt{2}a{)}^2}=\frac{1}{3-2\sqrt{2}}\Rightarrow r=3+2\sqrt{2}$

When $d=-\sqrt{2}a,$
$r=\frac{a^2}{(a+\sqrt{2}a{)}^2}=\frac{1}{3+2\sqrt{2}}\Rightarrow r=3-2\sqrt{2}$

Hence, the possible value(s) of $r=1,3\pm 2\sqrt{2}$