Find $ a_10 $ of a geometric progression if $ a_1 = 10 ~~ \text{and} ~~ r = -2 $.
To find $ a_{ 10 } $ we use formula
$$ \color{blue}{a_n = a_1 \cdot r^{n-1}}$$In this example we have $ a_1 = 10 ~~,~~ r = -2 ~~\text{and}~~ n = 10 $. After substituting these values to above formula, we obtain:
$$ \begin{aligned} a_n &= a_1 \cdot r^{n-1} \\ a_{ 10 } &= 10 \cdot \left( -2 \right)^{ 10 - 1} \\ a_{ 10 } &= 10 \cdot \left( -512 \right) \\ a_{ 10 } &= -5120 \end{aligned}$$The first few terms of this sequence are:
$$ 10, ~~~-20, ~~~40, ~~~-80 . . . $$