Question

For what values of $x$, the numbers $-\frac{2}{7},x,-\frac{7}{2}$ are in G.P.?

Answer 1

Given $-\frac{2}{7},x,-\frac{7}{2}\in$ G.P
$\Rightarrow x^2=-\frac{2}{7}\cdot -\frac{7}{2}=1$
$\Rightarrow x^2=1\therefore x=\pm 1$
Since if $a,b,c\in$ G.P then $b^2=ac$