## Algebra 2 (1st Edition)

$a_n=120 (1.5)^{n-1}$ and $a_7=1366. 875$
The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, $a_1=120; r=\dfrac{a_2}{a_1}=\dfrac{180}{120}=1.5$ Equation (1) gives: $a_n=120 (1.5)^{n-1}$ Plug in $n=7$ $a_7=120 (1.5)^{7-1} \approx 1366. 875$ Hence, $a_n=120 (1.5)^{n-1}$ and $a_7=1366. 875$