Question

If $x,y,z$ are three consecutive terms of a $GP$, where $4x>0$ and the common ratio is $r$, then the inequality $z+3x>4y$ 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 $r\in (3,\infty )$
C $r=\frac{24}{5}$
D $r=\frac{1}{2}$

Since $x,y,z\in$G.P with common ratio $r$
$\Rightarrow y=xr,z=x{r}^2$
Now given, $z+3x>4y$
$\Rightarrow x{r}^2+3x>4\left(xr\right)$
$\Rightarrow x(r^2+4r-3)>0$
$\Rightarrow (r-1)(r-3)>0$, since $x>0$
$\Rightarrow r\in (-\infty ,1)\cup (3,\infty )$
Hence all options are correct.