Find $ a_12 $ of a geometric progression if $ a_1 = 1 ~~ \text{and} ~~ r = 3 $.
To find $ a_{ 12 } $ we use formula
$$ \color{blue}{a_n = a_1 \cdot r^{n-1}}$$In this example we have $ a_1 = 1 ~~,~~ r = 3 ~~\text{and}~~ n = 12 $. After substituting these values to above formula, we obtain:
$$ \begin{aligned} a_n &= a_1 \cdot r^{n-1} \\ a_{ 12 } &= 1 \cdot 3^{ 12 - 1} \\ a_{ 12 } &= 1 \cdot 177147 \\ a_{ 12 } &= 177147 \end{aligned}$$The first few terms of this sequence are:
$$ 1, ~~~3, ~~~9, ~~~27 . . . $$