Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For n greater than 1, the area of nth smallest circle in square inches, A_n, is given by $$A_n=A_{n-1}+\left(2n-1\right)\pi.$$
What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?
$$A.\ 1$$
$$B.\ 1\frac{1}{2}$$
$$C.\ 2$$
$$D.\ 2\frac{1}{2}$$
$$E.\ 3$$
The OA is B.
I don't have clear this PS question,
Can I say that,
$$A_1=\pi,\ r_1=1$$
$$A_2=A_1+\ 3\pi=4\pi,\ r_2=2$$
$$A_3=A_2+\ 5\pi=9\pi,\ r_3=3$$
$$A_4=A_3+\ 7\pi=16\pi,\ r_4=4$$
Then, I stuck without ideas about how to continue with the solution.
I appreciate if any expert explain it for me. Thank you so much.
What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?
$$A.\ 1$$
$$B.\ 1\frac{1}{2}$$
$$C.\ 2$$
$$D.\ 2\frac{1}{2}$$
$$E.\ 3$$
The OA is B.
I don't have clear this PS question,
Can I say that,
$$A_1=\pi,\ r_1=1$$
$$A_2=A_1+\ 3\pi=4\pi,\ r_2=2$$
$$A_3=A_2+\ 5\pi=9\pi,\ r_3=3$$
$$A_4=A_3+\ 7\pi=16\pi,\ r_4=4$$
Then, I stuck without ideas about how to continue with the solution.
I appreciate if any expert explain it for me. Thank you so much.
