Find $ a_9 $ of a geometric progression if $ a_1 = 6 ~~ \text{and} ~~ r = -3 $.
To find $ a_{ 9 } $ we use formula
$$ \color{blue}{a_n = a_1 \cdot r^{n-1}}$$In this example we have $ a_1 = 6 ~~,~~ r = -3 ~~\text{and}~~ n = 9 $. After substituting these values to above formula, we obtain:
$$ \begin{aligned} a_n &= a_1 \cdot r^{n-1} \\ a_{ 9 } &= 6 \cdot \left( -3 \right)^{ 9 - 1} \\ a_{ 9 } &= 6 \cdot \left( 6561 \right) \\ a_{ 9 } &= 39366 \end{aligned}$$The first few terms of this sequence are:
$$ 6, ~~~-18, ~~~54, ~~~-162 . . . $$