Term and sum of a geometric progression

Statement

Given the geometric progression $2,\ 6,\ 18,\ 54,\ \dots$, find the tenth term and the sum of the first $10$ terms.

Solution

First term $a_1 = 2$, common ratio $q = 3$. The general term is $a_n = a_1\,q^{\,n-1}$: $a_{10} = 2\cdot 3^{9} = 2\cdot 19683 = \boxed{39366}.$ The sum of the first $n$ terms is $S_n = a_1\dfrac{q^{n}-1}{q-1}$: $S_{10} = 2\cdot\frac{3^{10}-1}{3-1} = 2\cdot\frac{59049-1}{2} = \boxed{59048}.$