## Algebra 2 (1st Edition)

$a_n=\dfrac{10}{9} 3^{n-1}$
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Given: $a_3=10$ and $a_6= 270$ $r^{6-3}=\dfrac{270}{10} \implies r=3$ Equation (1) gives: $a_3=a_1r^{3-1}$ Plug in $a_3=10$ $10=a_1 \times 3^2 \implies a_1=\dfrac{10}{9}$ This gives: $a_n=\dfrac{10}{9} 3^{n-1}$ Hence, $a_n=\dfrac{10}{9} 3^{n-1}$