## Algebra 2 (1st Edition)

$a_n=(2n-1) \pi$
The general formula for the nth term of an arithmetic series is given by $a_n= a_1+(n-1)d$ ...(1) Since $A= \pi r^2$ Now, we have $a_1=\pi (1)^2 =\pi \\a_2=\pi (2)^2 -\pi =3\pi \\a_3=\pi (3)^2 -(2)^2 \pi =5\pi ....$ Equation (1) gives: $a_n= a_1+(n-1)d$ and $a_n= \pi+(n-1) \times 2 \pi=(2n-1) \pi$ Hence, $a_n=(2n-1) \pi$