Answer
$1$
Work Step by Step
We know that $a_n=a_1r^{n-1}.$
where $r$=common ration (we can get this by dividing any two consecutive terms) and $a_1$= the first term
Here $a_1=1$ and $r=\frac{a_2}{a_1}=\dfrac{-1}{1}=-1$
Hence, here we have
$a_n=1(-1)^{n-1}.$
Therefore $a_{9}=1(-1)^{9-1}\\=1.$