Question

If $x,y,z$ are three consecutive terms of a $GP$, where $x>0$ and the common ratio is $r$, then the inequality $z+3x>3y$ holds for

• A. $r\in (-\infty ,1)$
• B. $r=\frac{24}{5}$
• C. $r\in (3,+\infty )$
• D. $r=\frac{1}{2}$

Answer 1

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

The correct options are
A $r\in (-\infty ,1)$
B $r=\frac{24}{5}$
C $r\in (3,+\infty )$
D $r=\frac{1}{2}$

$x>0$

$y=rx,z=r^{2}x$
$z+3x>3y$

$=>r^{2}x+3x>3rx$
$=>x(r^2-3r+3)>0$

$(r^2-3r+3)>0$
$(r-\frac{3}{2}{)}^2+\frac{3}{4}>0$

which is true for all R.

Hence, options A,B,C,D are correct.