Find $ a_10 $ of a geometric progression if $ a_1 = 3 ~~ \text{and} ~~ r = 5 $.
To find $ a_{ 10 } $ we use formula
$$ \color{blue}{a_n = a_1 \cdot r^{n-1}}$$In this example we have $ a_1 = 3 ~~,~~ r = 5 ~~\text{and}~~ n = 10 $. After substituting these values to above formula, we obtain:
$$ \begin{aligned} a_n &= a_1 \cdot r^{n-1} \\ a_{ 10 } &= 3 \cdot 5^{ 10 - 1} \\ a_{ 10 } &= 3 \cdot 1953125 \\ a_{ 10 } &= 5859375 \end{aligned}$$The first few terms of this sequence are:
$$ 3, ~~~15, ~~~75, ~~~375 . . . $$